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NE^WCOMB'S 

Mathematical  Course, 


Algebra  for  Schools,                            .         .         .         .  $i  20 

Key  to  Algebra  for  Schools,        ....  i  20 

Elements  of  Geometry,    .         .                  .         .         .  i  50 

The  Essentials  of  Trigonometry,        .                  .  i  25 

Plane  Geometry  and  Trigonometry,  with  Tables,  i  40 

Algebra  for  Colleges,   .                                    .  i  60 

Key  to  Algebra  for  Colleges,           .         .         .         .  I  60 

Plane  and  Spherical  Trigonometry,  with  Tables,  2  00 

Trigonometry  {separate), I  50 

Tables  {separate), i  40 

Elements  of  Analytic  Geometry,   .                 :        .  i  50 
Calculus  {in  preparation),        .         .          .         > 

Astronomy  (Newcomb  and  Holden),                      v  •  '2  -50 

The  same,  Briefer  Course,                   ...  i  40 


//£yi//?K  HQLl  &  CO.,  Publishers,  New  York. 


NEWGOMB'8    MATHEMATICAL    COURSE 


A 


SCHOOL   ALGEBRA 


BY 


SIMON  E-EWOOMB 

Professor  of  Mathematics  in  the  Johns  Hopkins  University 


NEW  YORK 
HENRY  HOLT  AND  COMPANY 

1887 


Copyright,  1882, 
HENRY  HOLT  &  OO. 


PEEFACE. 


The  guiding  rule  in  the  preparation  of  the  present  work 
has  been  to  present  the  pupil  with  but  one  new  idea  at  a  time, 
and,  by  examples  and  exercises,  to  insure  its  assimilation  be- 
fore passing  to  another.  This  system  is  carried  out  by  a 
minute  subdivision  of  all  the  algebraic  processes,  and  the 
end  kept  in  view  is  that  in  no  case  where  it  can  be  avoided 
shall  the  pupil  have  to  go  through  a  process  of  which  he  has 
not  previously  learned  all  the  separate  steps.  As  a  part  of 
the  same  plan,  definitions  are  so  far  as  practicable  brought  in 
only  as  they  are  wanted,  the  first  exercises  in  indicated  opera- 
tions are  performed  with  numbers,  and  the  pupil  is  set  to 
work  on  exercises  from  the  start. 

Correlative  with,  the  system  of  subdivision  is  that  of  ex- 
tending the  scope  of  the  exercises  so  as  to  include  not  only 
the  elementary  operations  of  algebra,  but  their  combinations 
and  applications.  It  is  hoped  that  the  wider  range  of  thought 
and  expression  in  which  the  pupil  is  thus  practised  will  be 
found  to  tell  in  his  subsequent  studies. 

As  another  part  of  the  same  general  plan  the  subject  has 
been  divided  into  three  separate  courses.  The  First  Course, 
which  extends  to  Simple  Equations,  is  intended  to  drill  the 
student  in  all  the  fundamental  processes  by  exercises  which 
are,  for  the  most  part,  of  the  simplest  character.  The  varied 
exercises  in  algebraic  expression  which  are  scattered  through 
this  course  form  a  feature  to  which  the  attention  of  instructors 
is  especially  solicited. 

In  the  Second  Course  the  processes  are  combined  and  the 
whole  subject  is  treated  on  a  higher  plane.  The  general 
arrangement  of  this  course  is  the  same  as  that  of  the  Elemen- 
tary Course  in  the  author's  College  Algebra.  But  the  exer- 
cises are  all  different,  and  greater  simplicity  of  treatment  is 
aimed  at.    This  course  terminates  witli  Quadratic  Equations. 

183977 


iv  PREFACE. 

The  Third  Course  consists  of  three  supplementary  chapters 
which,  however,  should  be  mastered  before  entering  college. 

An  attempt  has  been  made  to  treat  quadratic  equations 
with  such  fulness  as  to  avoid  the  usual  necessity  of  reviewing 
that  subject  after  entering  college. 

In  the  preparation  and  use  of  such  a  work  no  question  is 
more  difficult  than  that  of  the  extent  to  which  rigorous  de- 
monstrations of  the  rules  and  processes  shall  be  introduced. 
At  one  extreme  we  have  the  old  method,  in  which  the  teach- 
ing is  purely  mechanical;  at  the  other,  the  modern  demand 
that  nothing  be  taught  of  which  the  reason  is  not  fully  ex- 
plained to  and  understood  by  the  pupil.  The  latter  method 
should  of  course  be  preferred,  but  we  meet  the  insuperable 
difficulty  that  we  are  dealing  with  a  subject  of  which  the 
reasoning  cannot  be  understood  until  the  pupil  is  familiar 
with  it.  The  rule  adopted  in  the  present  case  has  been  to 
present  a  proof,  reason  or  explanation  wherever  it  was 
thought  it  could  be  clearly  mastered.  Most  teachers  will, 
however,  admit  that  long  explanations  of  any  kind  weary 
the  pupil  more  than  they  instruct  him,  and  that  the  best 
course  is  to  present  the  examples  and  exercises  in  such  a 
form  that  their  logical  correctness  shall  gradually  become 
evident  without  much  further  help. 

Were  the  author  to  make  a  suggestion  respecting  the 
system  of  teaching  to  be  adopted,  it  would  be  to  commence 
the  study  of  algebra  at  least  one  year  earlier  than  usual,  and 
to  devote  the  whole  of  that  year  to  the  first  course,  taking 
two  or  perhaps  three  short  lessons  a  week.  The  habit  of 
using  algebraic  symbols  in  working  and  thinking  would  thus 
V)ecome  established  before  taking  up  the  subject  in  its  more 
difficult  forms. 


TABLE  OF  CONTENTS, 


FIEST  COURSE. 

ELEMENTARY  PROCESSES  OF  ALGEBRA. 

Chapter  I.    Algebraic  Language. 

PAOB 

Reference  List  of  Algebraic  Signs  and  Symbols 2 

General  Definitions 3 

Algebraic  Signs 4 

Parentheses 6 

Sign  of  Multiplication 7 

"     "  Division 9 

Symbols  of  Quantity 11 

Powers  and  Exponents 13 

Coefficients 15 

Exercises  in  Algebraic  Expression 16 

Chapter  II.    Algebraic  Operations. 

Section  I.    Definitions 19 

11.    Addition  and  Subtraction 19 

III.  Multiplication 27 

IV.  Division 34 

Miscellaneous  Exercises  in  Expression 41 

Memoranda  for  Review 42 

Chapter  III.     Algebraic  Fractions. 

Section  I.    Multiplication  and  Division  of  Fractions 43 

II.     Reduction  of  Fractions 49 

III.  Aggregation  and  Dissection  of  Fractions 53 

IV.  Multiplication  and  Division  by  Fractions 57 

Miscellaneous  Exercises  in  Fractions 60 

Memoranda  for  Review  62 


Vi  CONTENTS. 

Chapter  IV.    Simple  Equations. 

PAGE 

Definitions .63 

Axioms 64 

Transposing  Terms 65 

Division  of  Equations ...  68 

Multiplication  of  Equations 71 

Problems  leading  to  Simple  Equations 73 

Memoranda  for  Review 80 


SECOND   COUESE. 
ALGEBRA  TO  QUADRATIC  EQUATIONS. 

Chapter  I.    Theory  op  Algebraic  Signs. 

Use  of  Positive  and  Negative  Signs, 83 

Algebraic  Addition 87 

Subtraction 89 

Rule  of  Signs  in  Multiplication 90 

"     "      "     '*  Division 91 

w     ««      «     '*  Fractions 92^ 

Chapter  II.  Operations  with  Compound  Expressions. 

Section  I.    Preliminary  Definitions  and  Principles 94 

Principles  of  Algebraic  Language 95 

II.     Clearing  of  Compound  Parentheses 96 

III.  Multiplication 97 

IV.  Division  and  Factoring 106 

V.     Fractions 116 

VI.     Substitution 124 

VII.     Highest  Common  Divisor 125 

Chapter  III.    Equations  of  the  First  Degree. 

Section  I.     Equations  with  One  Unknown  Quantity 131 

II.  "  "    Two  Unknown  Quantities 134 

III.  "  "    Three  or  More  Unknown  Quantities 139 

Problems  leading  to  Equations  of  the  First  Degree 143 

Interpretation  of  Negative  Results 149 

Memoranda  for  Review 152 

Chapter  IV.    Ratio  and  Proportion. 

Section  I.    Ratio 153 

II.     Proportion 158 

Problems  in  Ratio  and  Proportion 163 

Memoranda  for  Review 167 


CONTENTS.  Vii 
Chapter  V.    Powers  and  Roots. 

PAGE 

Section  I.    Powers  and  Roots  of  Monomials 168 

Fractional  Exponents 173 

Negative  Exponents 176 

II.     Powers  and  Roots  of  Polynomials 179 

Square  Root  of  a  Polynomial 184 

Square  Roots  of  Numbers . .  188 

III.     Operations  upon  Irrational  Expressions 192 

To  Complete  the  Square 207 

Memoranda  for  Review 209 

Chapter  VI.     Quadratic  Equations. 

Section  I.    Pure  Quadratic  and  Other  Equations 211 

II.     Complete  Quadratic  Equations 216 

III.  Equations  which  may  be  solved  like  Quadratics 224 

IV.  Factoring  Quadratic  Expressions 227 

V.     Solution  of  Irrational  Equations 232 

VI.     Quadratic  Equations  with  Two  Unknown  Quantities. . .  235 

VII.     Imaginary  Roots 240 

Memoranda  for  Review 243 


THIRD   COURSE. 

PROGRESSIONS,  VARIATION  AND  LOGARITHMS. 

Chapter  I.    Progressions. 

Section  I.     Arithmetical  Progression 247 

II.     Geometrical  Progression 253 

Limit  of  the  Sum  of  a  Progression 258 

Chapter  II.    Variation 263 

Chapter  III.    Logarithms 271 

Table  of  Four-Place  Logarithms 278 


APPENDIX. 
Supplementary  Exercises. 

Factoring 281 

Problems  in  Ratio  and  Proportion 284 

Quadratic  Equations 288 

Progressions t 291 


FIRST    COURSE. 


THE     ELEMENTARY     PROCESSES 
OF   ALGEBRA. 


EEFEEE^CE   LIST 
OF  ALGEBEAIC    SIGNS  AND   SYMBOLS. 


r=     RepresentSy  sign  that 


-f-  Plus,  sign  of  addition. 

—  Minus,  sign  of  subtraction. 

X  Times,  sign  of  multiplication. 

-ir  Divided  by,  sign  of  division. 

Is  to,  or  divided  hy,'  sign  of  ratio. 

: :  So  is,  sign  of  equality  of  ratios. 

:=  Equals,  sign  of  equality. 

'  two  expressions  are  identically 

equal, 
a  symbol  stands  for  an  expres- 
sion (§  109). 

>     Greater  than.     A  ^  B  means  A  is  greater  than  B. 
<     Less  than,  A  <  B  means  A  is  less  than  B. 

signs  of  aggregation,  indicate  that 
the  included  expression  is  to  be 
treated  as  a  single  quantity. 

|/    Radical,  indicates  a  root. 
"|/    nth.  root. 

Continued,  stand  for  any  number  of  unwritten 

letters  or  terms. 
Because. 

Therefore,  or  hence. 
oc     Infinity,  represents  a  quantity  increasing  beyond  all 
limits. 


(       )     Parentheses, 
— Vinculwn, 


SCHOOL  ALGEBRA. 


CHAPTER    I. 

THE  ALGEBRAIC    UNGUAGE. 


P\^' 


General  Definitions, 


1.  Definition.  Mathematics  is  the  science  which 
treats  of  the  relations  of  magnitudes. 

2.  Def.  A  magnitude  is  that  which  can  be  divided 
into  any  number  of  separate  parts. 

Example.  Length,  breadth,  time  and  weight  are  magni- 
tudes. 

3.  Def.  A  quantity  is  a  definite  portion  of  any 
magnitude. 

Example.  Any  number  of  feet,  miles,  degrees,  bushels, 
years  or  dollars  is  a  quantity. 

4.  In  arithmetic  and  algebra  quantities  are  ex- 
pressed by  means  of  aarotos?.    /l  )  ( i  v  vj     '  ,^  ^  To^ 

To  express  a  quantity  by  a  Tminhrr  tt  take  a  cer- 
tain portion  of  the  magnitude  as  a  unit,  and  state  how 
many  of  the  units  the  quantity  contains. 

Example.  We  may  express  the  length  of  a  string  by  say- 
ing that  it  is  7  feet  long.  This  means  that  we  take  a  certain 
length  called  a  foot  as  a  unit,  and  that  the  string  is  ec^ual  to 
7  of  these  units. 


4  .  ALGEBRAIC  LANGUAGE. 

Algebraic   Signs. 

5,  Sign  of  equality. 

—  read  equals.^  or  is  equal  to.,  is  the  sign  of  equal- 
ity. It  indicates  that  the  quantity  which  precedes  it 
is  equal  to  the  quantity  which  follows  it. 

G.  Signs  of  addition  and  subtraction. 

+  read  plus,  is  the  sign  of  addition.  It  indicates 
that  the  quantity  which  foUows  it  is  to  be  added  to 
the  quantity  which  precedes  it. 

Def,  The  number  resulting  from  the  addition  is 
called  the  sum. 

—  read  minus.,  is  the  sign  of  subtraction.  It  indi- 
cates that  the  quantity  which  follows  it  is  to  be  sub- 
tracted from  the  quantity  which  precedes  it. 

Def.  The  quantity  from  which  we  subtract  i15  the 
minuend. 

The  quantity  subtracted  is  the  subtrahend. 

The  result  is  the  remainder. 

Examples.     The  expression 

9  +  4  =  13 
means  4  added  to  9  makes  13. 

9-4  =  5 
means  4  subtracted  from  9  leaves  5. 

Remark.  Numbers  to  be  added  or  subtracted  may  be 
written  in  any  order.  Thus  —5  +  7  is  the  same  as  7  —  5, 
so  that  -  5  -f-  7  =  7  -  5  =  2. 

EXERCISES. 


1. 

15  +    7  =  what? 

2. 

15  -  7  =  what? 

3. 

-  7  +  15  =  what? 

4. 

—  8  +  8  =  what? 

7.  Positive  and  negative  quantities. 
Def    The  signs  4-  and  —  are  called  the  algebraic 
signs. 

The  sign  +  is  called  the  positive  sign. 
The  sign  —  is  called  the  negative  sign. 


ALGEBRAIC  SIGNS,  5 

Def.  Positive  quantities  are  those  which  have 
the  sign  +  before  them. 

Negative  quantities  are  those  which  are  preceded 
by  the  sign  — . 

8.  We  may  have  any  number  of  quantities  con- 
nected by  the  signs  +  and  — . 

Example.     The  expression 

9-2-4+5 
means  2  to  be  subtracted  from  9,  leaving  7;  then  4  more  to 
be  subtracted,  leaving  3;  then  5  to  be  added,  making  8. 

Hence  we  may  write 

9-2-4  +  5  + 12 -2  =  18. 

But  when  we  have  several  quantities  connected  by  the  signs 
+  and  —  it  is  generally  easiest  to  add  all  the  positive  (Quanti- 
ties and  all  the  negative  quantities  separately,  and  then  subtract 
the  sum  of  the  negative  from  the  sum  of  the  positive  quanti- 
ties. 

Example.     The  preceding  expression  is  calculated  thus: 
9-2 


5-4             26 

12-2           -  8 

26-8              18 

"We  add  9,  5,  and  12,  making  26.     Then  we  add  -2,-4,  and 

making  —  8.     Then 

.  36  -  8  =  18. 

EXERCISES. 

Compute  the  values  of  the  following  expressions*. 

1. 

14  _    3  _    8  +  22  +    17. 

2. 

41  -    9  -  10  -  11  +      1. 

3. 

12  -  13  +  14  -  15  +    16. 

4 

1+2+3+4-      9. 

6. 

1  +    3  +    6  +  10  +    15. 

6. 

14-    4  —    9  _  16  +    25. 

7. 

1  +    8  +  27  +  64  +  125. 

8. 

1  -    8  +  27  -  64  +  125. 

9. 

24  -  31  +    1  -    2  +    50. 

10. 

9  -|_  19  _|_  29  -  39  +    40. 

11. 

_  24  -  13  +    7  -    4  +  101. 

ALGEBRAIC  LANGUAGE. 


12. 

9-        5+      5-        9  +  14. 

13. 

-  32  -      23  +    50  +      32  +  23. 

14. 

117  _      13  -    31  -      17  -  56. 

15. 

1008  -  1008  -  500  +    650. 

16. 

1205  +  1336  -  296  -  1694. 

17. 

439  +    940  -  631  -    142. 

Parentheses, 

9.  When  combinations  of  numbers  are  enclosed  in 
parentheses  the  results  to  which  they  lead  are  to  be 
treated  as  if  they  were  single  numbers  or  quantities. 

Example.  19  —  (9  —  4)  means  that  the  quantity  9  —  4, 
that  is  5,  is  to  be  subtracted  from  19.     The  remainder  is  14. 

Therefore 

19  -  (9  -  4)  =  19  -  5  =  14. 
The  expression  12  -  (7  -  3)  +  (8  -2)  -  (9  -  6)  means 
that  7  —  3,  that  is  4,  is  to  be  subtracted  from  12;  then  8  —  2, 
that  is  6,  is  to  be  added;  then  9  —  6,  that  is  3,  is  to  be  sub- 
tracted. 

We  may  write  the  expression  thus: 
12  -  (7  -  3)  +  (8  -  2)  -  (9  -  6)  =  12  -  4+  6  -  3  =  11. 

EXERCISES. 

Compute  the  values  of  the  following  expressions: 

1.  12 -(11 -10). 

2.  17  -  (13  -  4). 

3.  17 -(13 +  4). 

4.  46 +  (19 -4). 
6.  46 -(19 -4). 

6.  13 +  (14 +  15). 

7.  27- (16-9-4). 

8.  41 -(31  +  1 -7). 

9.  (101  -  99)  +  (1  +  8  +  27). 

10.  (34  + 13  -  9)  -  (34  -  13  -  9). 

11.  1  +  3  +  6  -  (1  -  3  +  6). 

12.  (1  +  3  +  6)  -  1  -  3  +  6. 

13.  (9  -  8)  -  (8  -  7)  +  (7  -  6). 

14.  _  (9  _  8)  +  18  -  (33  -  17). 

15.  (74  -  69)  -  (8  +  3  -  7)  +  145. 


SIGI^'JS   OF  MVLTIFLICATION  7 

16.  (27  +  8)  -  (27  -  8)  4-  (1  +  2  -  3)  -  (1  -  2  -h  3). 

17.  -(ll-7)-l-(134-T-19-0)-27+(lH-2-3+4-5). 

18.  (136  -  48)  -f-  (49  -  1  -  35)  -  (10  -  7  +  13). 

19.  (7»-ll+13)-(7  -11-4-13)  +(13-11)  -  (11-13+4). 

20.  (746  -  614)  --  (42  -  18  -  17)  +  973-39-(973  -  39). 

21.  8  +  1- (8-  1). 

22.  8  +  2  -  (8  -  2). 

10.  Signs  of  Multiplication. 

X  read  multiplied  by,  or  times^  indicates  tliat  the 
numbers  between  which  it  stands  are  to  be  multiplied 
together. 

Example  1.  12  x  5  =  60  means  12  multiplied  by  5 
make  60. 

Ex.  2.  (11  -  -  3)  ;<  (4  +  2)  means  that  the  remamder, 
when  3  is  subtracted  from  11,  is  to  be  multiplied  by  the  sum 
4  +  2,  that  is  6.     Therefore 

(11  -  3)  X  (4  +  2)  =  8  X  6  =  48. 

Def.  The  quantity  to  be  multiplied  is  called  the 
multiplicand. 

The  number  by  which  it  is  multiplied  is  the  multi- 
plier. 

The  result  is  called  the  product. 

Multiplier  and  multiplicand  are  called  factors. 

11.  Omission  of  the  sign  x.  The  sign  X  is  gen- 
erally omitted,  and  when  two  quantities  are  written 
after  each  other  without  any  sign  between  them  it  in- 
dicates that  they  are  to  be  multiplied  together. 

Between  numbers  a  dot  may  be  inserted. 

The  reason  for  inserting  the  dot  is  that  such  a  product  as  3  X  5 
would  be  mistaken  for  25  (twenty-fire)  if  nothing  were  inserted. 

Examples.  2(5  +  3)  means  the  sum  of  5  and  3  multi- 
plied by  2.     Therefore 

2(5  +  3)  =r  2  .  8  =  16. 
3.7.2  means  3  times  7  times  2,  which  makes  42. 

12.  We  may  have  any  number  of  quantities  mul- 
tiplied together.     Any  two   of  them   are  then  to  be 


S  ALGEBtiAIC  LAMJL'AGE. 

multiplied  together ;  this  product  multiplied  by  the 
third,  this  product  multiplied  by  the  fourth,  this  pro- 
duct again  by  the  fifth,  etc. 

Examples.    2. '3. 4=2x3x4  =  6x4  =  24. 

3  .  5  .  2(1  +  3)  (1  4-  5)  =  3  .  5 .  2  .4 .  6  =  720. 

We   may   get   this  result  thus  :  3  .  5  =  15;    15  .  2  =  30  ; 
30  X  4  =  120;  120  X  6  =  720. 

The  multiphcations  may  be  performed  in  any  order  with- 
out changing  the  result. 

EXERCISES. 

Calculate  the  values  of  the  following  expressions: 

1.  4(9-2  +  3).    Ans.40. 

2.  (9  -  6)  (6  -  3).     Ans.  9. 

3.  (5  -2)  (10-  3).     Ans.  21. 

4.  (7  -  4)  (17  -  4).     Ans.  39. 

5.  3(13  -  7)  (19-  14).     Ans.  90. 

6.  (8  -  4)  4(9  -  5).     Ans.  64. 

7.  (1-2+3)  (4-5  +  6). 

8.  (1  +  2)  (2  +  3)  (3+4)  (4+5). 

9.  10(13  -  9)  (5  -  1). 

10.  17(17-1). 

11.  (13  +  7)  (13  -  7)  =  13  .  13  -  7  .  7. 

12.  (8  +  5)  (8  -  5)  =  8  .  8  -  5  .  5. 

13.  (19  -  3)  (19  -  3). 

14.  (14  +  4)  (14  -  4). 

15.  (1  +  9)  (1+10). 

16.  (3  +  4  +  5)  (3  +  4  -  5). 

17.  (3  +  4  +  5)  (3-4+5). 

18.  (3  -  4  +  5)  (3  +  4  -  5). 

19.  (25  +  10  +  4)  (5  -  2)  =  5  .  5  .  5  -  2  .  2  .  2.- 

20.  9(9  -  3)  -  8(8  -  3)  +  7(7  -  3). 

21.  (8  -  2)  (7  -  3)  -  (6  -  3)  (5  -  2). 

22.  (6.8-5.9)  (2.3.  4-4.  5). 

23.  (7.8-  2.12)  (5.6-4.7). 

24.  (2  .  3  .  4  -  20)  (4  .  5  .  6  -  7  .  5  .  3). 

25.  3(2.6-3.1)  (4.  5  -2.  8). 

26.  2(2  .  3  -  3  .  1)  (4  .  5  -  2  .  7)5. 


SIOJVS  OF  DIVISION.  9 

13.  Signs  of  Division. 

-i-  read  divided  by,  indicates  that  the  number 
which  precedes  it  is  to  be  divided  by  that  which 
follows  it. 

Examples.  15  -j-  3  =  5; 

19  -f-  7  =  ^  =  2f . 

Def.  The  quantity  to  be  divided  is  called  the 
dividend. 

That  by  which  it  is  divided  is  the  divisor. 

The  result  is  called  the  quotient. 

14.  Division  in  algebra  is  commonly  expressed  by 
writing  the  divisor  below  the  dividend  with  a  line  be- 
tween them,  thus  forming  a  common  fraction,  which 
fraction  will  be  the  quotient. 

Illustration.     Let  us  show  that  the  quotient  2  -i-  3  =  f . 

This  means  that  if  we  take  two  units  and  divide  the  sum 

of  them  into  three  equal  parts,  each  of  these  parts  will  be  f 

of  the  unit. 

I unit I unit 1 

lilililililil 
Let  us  draw  a  hne  as  above,  two  inches  in  length,  the 
inch  being  the  unit.     Divide  each  unit  into  three  equal  parts. 
Each  of  these  parts  will  then  be  \.     Since  the  two  units  have 
6  parts  in  all,  one  third  of  the  line  will  be  2  parts,  that  is  f . 

EXERCISES. 

1.  Show  in  the  same  way  that  if  we  divide  a  line  3  units 
long  into  5  equal  parts,  each  part  will  be  f  of  the  unit. 

We  first  divide  each  unit  into  5  parts,  making  15  fifths. 
Dividing  these  15  parts  by  5  we  shall  have  3  parts,  that  is 
|,  for  the  quotient. 

2.  Divide  a  line  5  units  long  into  3  parts,  and  show  that 
each  part  is  |  of  the  unit. 

We  divide  each  unit  into  thirds,  making  15  thirds  in  all. 

3.  Divide  a  line  7  units  long  into  4  parts,  and  show  that 
each  part  is  J  of  the  unit. 

.    9  +  20-2_27_ 
7-4       ~  "3  ~    • 


10  ALGUBBAIG  LANGUAGE. 

2.3(6  +  8)         2.3.14_ 
(3  +  4)  (5  -  3)  7.3 

3  +  12  -  (9  -  7)  _  15  -  2  _  13  _  „, 
^-  2"."  3 ~6~  -  T  -  ^• 

Note,  We  obtain  this  result  by  multiplying  the  numerator  of  the 
fraction  by  the  multiplier  7.  It  is  explained  in  arithmetic  that  we 
multiply  a  fraction  by  multiplying  its  numerator. 

13-4  +  9 


8. 


10.   9  X 
11 


-(1  +  2)2  • 
14  -  (14  -  4)  +  17 
1+2+4        • 
(2  +  1)  (2  +  1) 


(1  +  2)  (1  +  2-3  +  4)- 

(8  +  1)  (1  +  8)8  +  1 
8-1 


12    2        l  +  (l  +  2)2  +  (l  +  2  +  3)3 
^-^^  '^  •  2  (4  -  3  +  2  -  1)  +  3  (3  -  2  +  1)* 
j3    1.3.5.7 

14. 
15. 
16. 
17. 
18. 
19. 


2. 

4.6.8* 

l  +  7.4(7  +  4)(7- 

4) 

^  (4  +  7)  +  7 

8. 

7.6.5'      8.7.6 

.5.4 

1. 

2.3.4  •   1.2.3 

.4.5' 

(2 

.  2  +  3)  (3  .  3  -  2) 

7x7 

13 

.13  +  31 

2, 

.5.5.2' 

(1 

.  3)  +  (3  .  5)  +  (5  . 

7) 

(2 

.  4)  +  (4  .  6)  +  (6  . 

8)- 

(1 

.  1  .  1)  +  (2  .  2  .  2)  +  (3  . 

3  .  ;3)  +  (4  .  4  .  4) 


(1  +  2  +  3  +  4)  X  (1  +  2  +  3  +  4) 
2.2.4.4.6.6 


^^'  ^^1.3.3.5.5.7* 


REDUCTION  TO  NUMBERS.  11 

Symbols  of  Quantity. 

15.  In  algebra  quantities  are  represented  by  the 
letters  of  the  alj^habet. 

Def.  The  letters  used  to  represent  quantities  are 
called  symbols. 

16.  A  quantity  may  be  represented  by  any  symbol  we 
choose.  The  symbol  may  then  be  regarded  as  a  7imne  for  the 
quantity. 

Example.  We  may  draw  a  straight  line  and  call  it  a. 
The  letter  a  is  then  the  name  of  that  line.  But  we  might 
also  have  called  it  h,  c,  Xy  or  any  other  name  we  choose.  The 
only  restriction  is  that  two  quantities  must  not  have  the  same 
symbol  or  name  in  the  same  question. 

17.  Def.  The  value  of  a  symbol  is  the  number  or 
quantity  which  it  represents. 

Example.  In  the  preceding  example  the  length  of  the 
line  is  the  value  of  the  symbol  a. 

If  I  have  the  number  275  and  call  it  n,  then  275  is  the 
value  of  n. 

Reduction  of  Algebraic  Expressions  to 
Numbers. 

18.  Def.  An  algebraic  expression  is  any  com- 
bination of  algebraic  symbols. 

Rule.  To  reduce  an  algebraic  expression  to  mimbers  we 
put,  171  place  of  the  symbols,  the  numbers  tvhich  they  represent, 
and  perform  the  operations  indicated. 

Example.     Find  the  value  of  the  following  expression, 
en  —  ab 


supposmg 


3n- 

-6c* 

a  =  3, 

g  =  i^> 

6  =  6, 

n  =  15, 

6- =7, 

m  =  15. 

12 


ALGEBBAIC  LANQUAOE. 


We  do  this  as  follows: 
c/i  =  7  .  15  :=  105; 
a^>  =  3  .    G  =    18; 
hence  en  —  ab  —    87: 


3;i  =  3  .  15  =  45 

66'  =  6  .    7  =  42 

hence    3?^  —  6c  =    3 

and  then  V-  =  ^9- 


Ans. 


EXERCISES. 

Using  the  above  values  of  the  symbols  «,  h,  c,  g  and  n, 
compute  the  values  of: 

2.  a  -\-h  -\-  c  —  g. 
4.  gc  +  «&. 
6.  an  —{a  -{-h-{-  c). 
8.  ^7j  -\-  eg  —  be  -\-  ab, 
10.  2aJc  -  3^^  +  6a. 
bn  -|-  c^ 


1.    rt  +  ^  +  6'. 

3.  n  —  g  —  a. 
5.  ^c  —  «Z>. 
7.  abe  —  c(/. 
9.  3«^  —  4:be  -\-  bcgn, 
en  +  ab 


11. 
13. 

15. 


en  —  ab' 
aaa  +  bb 


9<^ 


2ng 
a^  b 


If  a  =  1,  ^ 
values  of: 
17.  ab\  abe;  abed;  abede. 

19.  ab  [1  +  c  +  cd  (1  +  e)] 

b  —  a      a  —  e 
23.  ^«  +  ^>(c  +  ^)(e+/). 

25.  {a-\-b-{-e-^  d)e^f. 


12. 
14. 

16. 


Z'/j  —  eg' 

Saaaa  —  2bbb  +  aabb 

aa  -\-  bb  —  9 
_gn_      n 
a  -\-b      a 


2,  c  =  3^  ^  =  4,  e  =  5,  etc.,  then  find  the 


18.  ab  +  abc  +  abed  +  «Jcc?e. 

22.  («  -{-b)(e  +  d)  {e  -f/). 
24.   {a-^b)e-\-d(e-{-f). 
26. 


(<?  +  ^)  ^  +  (^  +  c)g 
(Z*  -\-c)e^{c-^d)d-  V 


27.  4/(/-l). 


28. 


abe 


30.   (/-.)( 


def 
d){d-e){e-^d^e^f). 
6d  -  be  ^.    2«  +  Z> 


29. 


«ce 


4^  -  7«      5^  -  Qe 

33.  ^{1  +  ^  [2  4-^(3+'/)] 


32. 


2<^  +  « 


hdf 

\\^c{\-\-d)\. 


POWERS  AND   EXPONENTS.  13 

34.  Find  the  values  of  the  following  expression,  for  w  =  0, 
then  for  n  =  1,  then  for  n  =  2,  etc.,  to  n  =  7: 

n{n  -{-!)  {n  ^  2) 
6" 
For  w  =  0,  the  expr.  =0;  for  w  =  1,  the  expr.  =1;  iov  n  =  2, 
the  expr.  =  4,  etc. 

When  n  =  0,  the  expression  =  0, 
"  n  =  1,  "  "  =  1. 
"  n  =  2,  "  ''  =  4. 
"  n  =  3,  "  "  =  10. 
"  ?i  =  4,  "  "  =  20. 
-  w  =  5,  "  "  =  35. 
"  n  =  6,  "  "  =  56. 
"      n  =  7,        "        "        =  84. 

In  like  manner 'find  the  values  of: 

35.  in{7i  +  1),  for  7i  ^  0,  1,  2,  3,  4,  and  5. 

36.  ^n(n  +  1)  (j^''>i  +1)^  foi"  same  values. 

37.  iw  .  7i('/i  -|-  1)  (u  4-  1),  foi'  same  values. 

38.  IT — ; — -,  for  same  values. 
Zn-\-  1 

39.  Can  you  show  that  the  quantity  S  -\-  n-{-  (S  ~  n)  is 
the  same  for  all  values  of  ^  ? 


Powers  and  Exponents. 

19.  Bef.  A  power  of  a  quantity  is  the  product 
obtained  by  taking  the  quantity  as  a  factor  any  num- 
ber of  times. 

Def.  The  degree  of  a  power  means  the  number 
of  times  the  quantity  is  taken  as  a  factor. 

If  a  number  is  to  be  raised  to  a  power,  the  result  may,  in 
accordance  with  the  rule  for  multiplication,  be  expressed  by 
writing  the  number  the  required  number  of  times. 

Examples.     5.5  =  25  is  the  second  power  of  5. 
bbi  is  the  third  power  of  b. 
xxxxx  is  the  fifth  power  of  x. 

To  save  repetition,  the  number  whose  power  is  to  be  ex- 
pressed is  written  only  once,  and  the  number  of  times  it  is 
taken  as  a  factor  is  indicated  by  small  figures  written  after 
and  above  it. 


14  ALQEBBAIC  LANGUAGE. 

Examples.     Instead  of    5.5  we  write  5'. 

aa  ''       ''     a\ 
"       "  UlU  ''      "    v. 
Also,  (1  -f  3)^  means  (1  +  3)(1  +  3)(1  +  3)  =  4 .  4 .  4  =  64. 
(2.3y        ''      2.3.2.3  =  6.6  =  36. 

20.  Bef.    The  number  wMcli  indicates  a  power  is 
called  the  exponent  of  that  power. 

EXERCISES. 

Find  the  values  of: 
1.   2^;         4^;         2^         3';         5^^;         4^ 

,3.   2^3^;  (1-1-2)^(2  +  3); 

(l+2)(2  +  3)'[(l  +  2)(2  +  3)p. 

4.   l-h4^:         (1  +  4)^;         (7-5)'-^;         r_5^-*; 
(10  -  7)  (9  -  6)  (8  -  5)  =  3^ 

(10+5)^  10  +  5^  10'  +  2  .  5  .  10  +  5' 

^-  5         '  5      '  5  • 

6.   1  +  2  +  3^         1  +  (2  +  3)=^;         (1  +  2  +  3)^ 

^    r  + 2^  4-3^4- 4^ +  5^ 

8, 

3^-1 

10.  —    „    ^^r   -;         7.2^;         (7.2)^;         7\  2. 

(4 +  7)' +  11(7 +  4)- -(13 -2) 
(19  -  9)=^  +  2^  -  5^  -  7 

12.  (6  -  5)^  +  (6  -  4)^  +  (6  -  3)« 

+  (6 -2)^ +  (6-1)^=^^^)-'. 

13.  7(8-1)  (9 -2)  (10 -3)  (11 -4)^ 

U.   87  -  3  .  5';         (87  -  3  .  5)';         87  -  3\  5. 


(1  +  2  +  3  +  4  +  5)' 

2    • 

1  +  2  +  2'^  +  2^  +  2^ 

'  +  3' 

2^-1 

1+3  +  3^^  +  3^  +  3* 

+  3' 

COEFFICIENTS.  15 

Write  the  following  expressions  with  exponents: 
15.  mmm.  16.  ppppp. 

17.  (a  +  b)  {a  -\-h){a^  h),      18.  {a  -  h)  (a  ^  b)  {a  +  b). 
19.  xxyyzzyzz.  20.  nn{7n  +  w)  (m  +  ^). 

21.   (jo  +  ^)  (g  -\-p)rrr.  22.  jfif(:?;  -  ?/)  {^  -y){^-  y)- 

23.  a^a^a;  (a  -\- x)  (a -\-  x).        24.  ^/^y^  (/  +  ^)  (^  +»> 
25.   (a  +  J  +  c)  (Z>  +  c  +  flj)  (c  +  fl^  +  b). 

Find  the  values  of  the  following  expressions  when  «  =  5, 
b  =  S,  m  =  2,  71  =  S: 


26. 

{b  -  aym\ 

27. 

(b  -  «)- 

28. 

(b  -  my-\ 

29. 

(b  -  nr- 

30. 

(a  -  ny\ 

31. 

(771  +  7l)^ 

(a  +  7n)"' 

32. 

(b  -  my 

33. 

(a-m)^' 

'         b"" 

Coefficients. 

21.  Bef.    The  coefficient  of  an  algebraic  symbol 
is  any  number  which  multiplies  it. 

Example.     In  the  expression 

3x  -4:y  -\-  \%z, 
3  is  the  coefficient  of  a:;  4  is  the  coefficient  of  y;  12  is  the  co- 
efficient of  z. 

Use  of  Coefficients.     If  we  call  this  line  a,  a 

then  this  line  =  2«, — go 

this  line  =  3ff,  ■ — 3^^^ 


and  this  line  =  4^. 


Aa 


Remark.     Compare  this  definition  of  coefficient  with  that  of  expo- 
nent of  a  power. 

a-\-a-\-a-\-a  —  ^a,  and  4  is  here  the  coefficient; 
also,  a  X  a  X  fl^  X  «  =  a\  and  4  is  here  the  exponent. 
22.  A  coefficient  may  be  an  algebraic  symbol  or  a 
product  of  several  symbols. 

Any  quantity  may  be  supposed  to  have  the  coeffi- 
cient 1. 


16  ALGEBRAIC  LANGUAGE. 

Examples.  In  the  expression  mx,  rn  is  the  coefficient  of  x. 
In  the  expression  %7nax, 

2ma  is  the  coefficient  of  x; 

2m  is  the  coefficient  of  ax; 
and  2  is  the  coefficient  of  max, 

EXERCISES. 

In  the  expression 

2amx  +  bcyz  +  mpqr. 
What  is  the  coefficient  of  a;? 
What  is  the  coefficient  of  mx^ 
What  is  the  coefficient  of  2;? 
What  is  the  coefficient  of  yz? 
What  is  the  coefficient  ot  mp? 


23.    Exercises  in  Algebraic  Expression. 

Note.  The  object  of  the  following  exercises  is  to  practise  the  student 
in  algebraic  expression.  The  answers  are  therefore  to  be  expressed  or 
indicated  in  the  manner  of  algebra,  instead  of  being  calculated 

1.  How  many  dollars  will  7  knives  cost  at  S2  each? 

To  get  the  cost  we  must  multiply  2  by  7;  this  is  done  algebraically 
by  writing  7.2;  therefore  the  required  cost  is  7  .  2  dollars. 

2.  What  will  7  pounds  of  tea  cost  at  $2  per  pound? 

3.  If  I  buy  9  sheep  at  15  each,  and  15  turkeys  at  $2  eacli, 
what  is  the  total  cost? 

4.  A  man  had  $15,  and  spent  $3;  how  many  dollars  had 
he  left? 

5.  If  a  boy  has  a  cents  in  one  pocket  and  b  cents  in 
another,  how  many  has  he  in  both?  Ans.  a  -{-  h. 

6.  A  man  had  x  dollars,  and  paid  out  y  dollars;  how 
many  had  he  left? 

7.  If  I  buy  7  cows  at  $45  each,  and  3  horses  at  $175, 
express  the  total  cost?  Ans.  $(7  .  45  +  3  .  175). 

8.  A  huckster  sold  a  chickens  at  x  cents  each,  and  lost 
m  cents  on  his  way  home;  how  many  cents  had  he  left? 

9.  A  train  having  to  run  a  distance  of  m  miles,  went  for 
2  hours  at  the  rate  of  30  miles  an  hour;  how  much  farther 
had  it  to  go? 


EXERCISES  m  ALGEBRAIC  EXPRESSION,  17 

10.  A  man  bought  two  pounds  of  tea  for  x  cents;  how- 
much  would  one  pound  cost?  ,         x 

Ans.    -  cents. 

11.  If  8  pounds  of  tea  cost  %x,  what  is  the  cost  per  pound? 

Ans.   %-. 

u 

12.  Three  men  took  dinner  at  a  hotel  at  a  cost  of  m  dol- 
lars for  all  three;  how  much  had  each  to  pay? 

13.  One  boy  has  w  cents  and  another  n  cents;  if  they 
make  an  equal  division  of  the  money,  how  much  will  each 
have? 

14.  A  boy  had  h  cents,  and  out  of  them  gave  c  cents  to 
one  person  and  d  cents  to  another;  how  many  had  he  left? 

15.  A  quantity  m  is  to  be  multiplied  by  another  quantity 
/^;  express  the  product. 

16.  What  will  a  -\-  h  sheep  cost  at  x  dollars  apiece? 

17.  What  will  m  -f-  'n  pounds  of  beef  cost  at  x  -\-  y  cents 
a  pound? 

18.  A  man  bought  j(;  dollars'  worth  of  flour  at  q  dollars  a 
barrel;  how  many  barrels  were  there? 

19.  How  many  cents  are  there  in  x  dollars?  How  many 
dollars  in  x  cents? 

20.  A  man  had  x  dollars  in  one  pocket  and  x  cents  in 
another;  how  many  dollars  had  he  in  both?  How  many  cents? 

21.  A  man  went  out  with  k  dollars  in  his  pockets,  and 
paid  out  m  cents;  how  many  cents  had  he  left? 

22.  What  will  be  the  value  of  7i  houses  at  x  dollars  apiece? 

23.  How  many  dollars  will  m  pounds  of  beef  cost  at  .^ 
cents  per  pound? 

24.  A  man  bought  from  his  grocer  a  pounds  of  tea  at  .r 
cents  per  pound,  and  b  pounds  of  sugar  at  y  cents  i)er  pound, 
and  c  pounds  of  coffee  at  z  cents  per  pound;  how  many  cents 
would  the  bill  amount  to?     How  many  dollars? 

25.  A  man  bought/  pounds  of  flour  at  m  cents  per  pound, 
and  handed  the  grocer  an  x  dollar  bill  to  be  changed;  how 
many  cents  ought  he -to  receive  in  change?  How  many 
dollars? 

26.  Of  3  boys  one  had  n  cents  in  each  of  his  two  pockets, 
another  had  p  cents  in   each  of  his  three  i)0('kots,  and  the 


1 8  AL  a  EBUA  J  L '  L.  I  ^'G  UA  GE. 

third  had  q  cents  ;   if  they  make  an  equal  division  of  the 
money,  how  much  would  each  have? 

27.  A  huckster  had  3  large  baskets,  each  containing  p 
apples,  and  7  small  baskets,  each  containing  q  apples.  He 
divided  his  ai)ples  equally  among  the  10  baskets;  how  many 
were  there  in  each  basket? 

28.  The  sum  of  the  quantities  17  and  29  is  to  be  divided  by 
the  sum  of  the  quantities  5  and  3;  what  will  be  the  quotient? 

29.  Tlie  sum  of  the  quantities  a  and  h  is  to  be  divided  by 
the  sum  of  the  quantities  m  and  n-,  what  will  be  the  quotient? 

30.  A  man  left  to  his  children  a  bo'nds  worth  x  dollars 
each  and  s  acres  of  land  worth  y  dollars  an  acre;  but  he  owed 
m  dollars  to  each  of  q  creditors;  what  was  the  value  of  the 
estate? 

31.  Two  numbers  x  and  y  are  to  be  added  together,  their 

sum  multiplied  by  5,  and  the  product  divided  by  «  +  ^;  ex- 

s(x  +  y) 
press  the  (quotient.  Ans.  ,    .    . 

32.  The  sum  of  the  numbers  p  and  q  is  to  be  divided  by 
the  sum  of  the  numbers  a  and  h,  forming  one  quotient,  and 
the  difference  of  the  numbers  jt?  and  q  is  to  be  divided  by  the 
difference  of  the  numbers  a  and  b,  forming  another  quotient; 
express  the  sum  of  the  two  quotients. 

33.  In  the  preceding  exercise  express  the  product  result- 
ing from  multiplying  the  sum  of  the  quotients  by  m. 

34.  The  quotient  of  x  divided  by  a  is  to  be  subtracted 
from  the  quotient  of  y  divided  by  h,  and  the  remainder  mul- 
tiplied by  the  sum  of  x  and  y  divided  by  the  difference  be- 
tween/'and  _?/;  express  these  operations  in  algebraic  language. 

35.  'I'he  number  x  is  to  be  increased  by  n,  the  sum  is  to 
multiplied  by  a  +  h,  and  q  is  to  be  added  to  the  product,  and 
the  sum  is  to  be  divided  by  rs',  express  the  result. 

36.  The  quotient  of  x  divided  by  y  is  to  be  divided  by  the 
<|Uotient  of  a  divided  by  h. 

37.  The  quotient  of  x  divided  by  y  is  to  be  added  to  the 
quotient  of  a  divided  by  h,  and  the  sum  is  to  be  divided  by 
the  sum  of  m  and  n. 

38.  X  -\-  y  houses  each  had  a  -\-  h  rooms,  and  each  room 
w  -\-  n  pieces  of  furniture;  how  many  pieces  were  there  in  all? 


ALGEBRAIC  OPERATIONS.  1^ 

CHAPTER    II. 
ALGEBRAIC     OPERATIONS. 

Section  I.     Definitions. 

34.  Term.  The  terms  of  an  expression  are  the 
parts  connected  by  the  signs  +  and  — . 

Example.  In  the  expression  a-\-2b  —  3c  the  parts  a, 
2b  and  3c  are  the  terms,  there  being  three  terms  m  all. 

Algebraic  expressions  are  divided  into  monomials 
and  polynomials. 

A  monomial  consists  of  a  single  term. 

A  polynomial  consists  of  more  than  one  term. 

A  binomial  is  a  polynomial  of  two  terms. 

A  trinomial  is  a  polynomial  of  three  terms. 

Examples.     The  expression  l^^abx  is  a  monomial. 

%ab  +  omxy  is  a  binomial. 

I  -\-  m  -\-  n  and  x^  -j-  2axy  +  y^  are  trinomials. 

Factor.  When  any  number  of  quantities  are  multi- 
plied to  form  a  product,  each  of  them  is  called  a 
factor  of  the  product. 

Equation.  An  equation  is  composed  of  two  equal 
expressions  with  the  symbol  =  between  them. 

Example.     7  +  5  =  15  —  3  is  an  equation. 

Lilie  Terms.  Like  terms  are  those  which  contain 
identical  symbols  and  differ  only  in  their  numerical 
coefficients. 

Example,     a,  2a  and  Ha  are  like  terms; 
like;  9a:-}- 12a;  are  also  like;  but  l^pq  and  6qr  are  nnlike. 


Sectiot^  IL     Addition  and  Subteaction. 

Addition  of  Positive  Quantities. 

25.  The  addition  of  any  number  of  terms  may  be 
indicated  by  writing  them  down  with  the  sign  -f-  be- 
tween them. 


*20  ALQEBBAIC  OPERATIONS. 

Example.  The  sum  of  the  quantities  2x,  A:ax,  6b  and 
bbx  may  be  written 

2x  +  4:ax  +  5^  +  6bx. 

If  tlie  terms  are  all  U7ilike,  this  is  the  only  way  of  express- 
ing addition  of  algebraic  quantities. 

26.  Rule  for  Like  Positive  Terms.  If  like  positive 
terms  are  to  be  added,  take  the  swn  of  all  the  coefficients  of 
the  symbol  and  affix  the  symbol  to  their  sum. 

In  this  addition  a  symbol  without  a  coefficient  must  be 
considered  as  having  the  coefficient  1. 

Example  1.  If  we  have  to  add  together  the  quantities  x, 
HXy  4tXy  %x,  12x,  we  proceed  thus: 

Ix 
2x 

4:X 

Sx 


Sum  =  27:*; 

Ex.  2.     x-{-x-^x-\-x-{-x  =  5x. 
Ex.  3.     am  -\-  2a7n  +  3am  =  6am. 


EXERCISES. 

Add  up  the  quantities: 
1.  9a-{-7a-\-6a-\-Sa.  2.  x -\- 2x  +  Sx -}- ^x. 

3.   2y  -\-dy  +  by.  4.   "dab  +  ^ab  +  ab. 

5.  ab  -\-  ab  -\-  ab  -\-  ab. 

When  there  are  several  sets  of  similar  terms,  add 
each  set,  and  connect  the  sums  by  the  sign  +• 

Example.     Add 

a  +  2x  +  3«  4-  lab  -\-  6x  -{- hab  -\-  x  -\-  12a. 

Work: 
We  pick  out  all  the  symbols  a  and  write 

them  under  each  other  with  their  coeflBcients; 

then  all  the  symbols  «;  then  all  the  products  a6. 

We  then  add  the  coefficients,  thus  formmg  the 

s^"°-  Sum  =  lea  +  92:  -f-  12rh 


-la-{-2x-{- 

tab 

3a-{-  6x  -\- 

bah 

12a  +  Ix 

ADDITION.  21 

6.  Add  lU  +  4^  +  l^abc  7.       ah  +    'de 

a-\-%b  -\-  mabc  22ab  -\- I2e -^  d 

Sa  -{-  6b  -\-  12abc  29ab  +      e 

8.  {x  +  !/)  +  n^  +  y)  +  H^  +  y)'       Ans.  16(0^  +  y). 

9.  (a  -'  ^>)  +  2{a  -  b)  +  'S(a  -  b), 

10.  (rn  —  j»)  +  O^i  —  P)  +  3(m  —  jo)  +  24:(m  —  p). 

11.  3(m  —  /i)^  -j-  5(w  —  ^)^  +  7(/?i  —  ^O^'- 

12.  If  Thomas  has  x  dollars,  John  3  times  as  many  as 
Thomas,  and  James  as  many  as  John  and  Thomas  together, 
express  the  number  all  three  have  in  the  language  of  algebra. 

13.  If  a  man  owes  his  grocer  b  dollars,  his  tailor  c  dollars, 
his  shoemaker  4  times  as  much  as  his  grocer,  and  his  butcher 
twice  as  much  as  his  shoemaker,  what  is  his  total  indebted- 


ness 


Addition  of  Negative  Quantities. 

27.  Let  us  have  to  add  T  —  4,  that  is  3,  to  10.  This 
is  the  same  thing  as  adding  7  and  subtracting  4,  because 
10  +  7  —  4  is  the  same  as  10  +  3,  namely,  13. 

Whatsoever  numbers  a,  b  and  e  represent,  if  we  add  b  —  c 
to  a  the  sum  will  be  a  -\-  b  —  c. 

Hence 

Rule.  //"  negative  qtianfities  are  among  those  to  be  added, 
they  are  to  be  subtracted  from  the  sum  of  the  others. 

Algebraic  addition  therefore  means  something  more  than 
addition  in  arithmetic  or  arithmetical  addition,  because  it 
may  include  subtraction. 

28,  Def.  Algebraic  addition  means  the  combi- 
nation of  quantities  according  to  their  algebraic  signs, 
the  positive  ones  being  added,  and  the  negative  ones 
subtracted. 

Bef.  The  result  of  algebraic  addition  is  called  the 
algebraic  sum. 

Example.     The  algebraic  sum  of  ~  2—3  +  15  —  6  is  4. 

Def.  Numerical  addition  and  numerical  sub- 
traction mean  addition  and  subtraction  as  in  arith- 
metic, without  regard  to  the  algebraic  signs  of  the 
quantities.  /^^^5>v 


22  ALGEBBAIC  OPERATIONS. 

29.  KuLE  FOR  Algebraic  Additioi^.  Take  the  mm  of 
all  the  positive  terms  and  the  sum  of  all  the  negative  terms. 
Subtract  the  less  siom  fro7n  the  greater  and,  if  the  negative 
Slim  is  the  greater,  lurite  the  sign  —  before  the  difference. 

Remark.    If  the  positive  and  negative  sums  are  equal,  the  total  sum 

will  be  zero. 

Example  1.     4  -  2  +  3  -  8  =  +  7  -  10  =  -  3. 

Here  the  sum  of  the  positive  quantities  is  7  and  of  the  negative  ones 
10.  In  arithmetic  we  cannot  subtract  10  from  7,  but  in  algebra  we  ex- 
press this  subtraction  by  taking  7  from  10  and  writing  —  before  the 
difference  to  indicate  that  tlie  subtractive  quantities  are  the  greater. 

Ex.  2.  Add    4.T  -h  %ay  -  3z  Ex.  3.  Add     7x  -  5y  +  4 

3x  —  4:a?j  —    z  —  2x  —  dg  —  S 

2x  —  dag  -{-4:Z  —  5a;  +  8^  +  4 

Sum,  9x  —  6ag  Sum,      0        0        0 

EXERCISES. 

Add: 

1.  dx-4:X-^a-\-  Tx  +  5x  -i-2b-\-3a-  7.?;  -  6a. 

2.  26?^  —  2y  -^  (jg  -\-  Sag  —  Ig  -\-  lOxg  +  6ag  —  5xg. 

3.  8^-  +  9  -  3//  +  :x  -12-\-Sg  -  2x  -  3-  7g. 

4.  7?/  +  6?/  +  3rt  -  9//  -  2g  ~  ba -{-  2g  -  4^  +  2a. 

5.  9x  -\-  axg  -\-  3//  —  6x  —  2xg  -}-  6g. 

6.  9x  —  3xg  —  l^x  +  ^xg  —  \x  —  Qx. 

7.  4a  -  2b,     Oft  -  bb,     8a  —  lib,     a  +  75. 

8.  4.^•  —  dg,     3x  —  bg,     —  x  -\-  g,     —  ^x  +  4?/. 

9.  5a  +  35  +  c,     3a  +  35  +  3c,  a  +  35  +  be. 

10.  3X  +  2g  —  z,     2x—2g-\-  2z,     —  x -]- 2g -i- 3z. 

11.  7^f  _  4j  _|_  c,  6a  +  35  —  be,     —  12a  +  4c. 

12.  .t-  — 4a +  5,  32;-f25,  a  —  .t  —  55. 

13.  a  +  5  —  c,  5  +  6'  —  a,  c  +  a  —  5,  a  +  5  —  c. 

14.  a  4-  25  +  3c,     2a  —  b  —  2c,     b  —  a  —  c,     c  —  a  —  h. 

15.  ^_25+3c-4^,  35-4c+5fZ-2a,  5c-6^+3a--45. 

16.  a  -  25  +  c,  2a  -  35  -  4c,  3c  -  4^. 

17.  2x  +  3,  5^;  +  7,  13:c  +1,  ^  -  4. 

18.  ax-\-  5,  ca;  +  d,     ex  -\-f 

19.  m«/  -\^n,     bag  —  n,     2n  —  lag. 

20.  a+54-c,  a  +  5  —  c,  a  —  5  +  c,  — a+5  +  «» 

21.  x^^2xg^g\     x'-2xg^y\     2x^-2y\ 

22.  x^'-y^     g''-z\     z^-x\ 


SUBTRACTION.  23 

23.  llax  —  19hy     13a;  -  c,     lib  —  a-\-  c,     Hb  -  lOax. 

24.  7«  -  85  +  9c,     9b -7c-  Sa,     9a  -  n  ^  Sc. 
26.  a'+l,     a'-\-l,     a  +  1,     1  -  a,     1  -  a\ 

26.  a  -\-m,     n  -\-%a,    p—  3a,     Aa  —  q. 

27.  axi/  4-  P^^  +  '^y  +  ^>  4/?2:*  —  3«a;?/  +  9X;  —  Smy, 
l^axy  —  5^.r^  —  k. 

28.  m  —  2/i,  —  2m  +  3/i  —  4/?,  3w  —  4?^  +  ^j^  —  6§'? 
—  4m  +5^  —  6jo  +  7^  —  8r. 

29.  «:?;'+  %'—  a'5%     ax"  —  by""  +  tt'Z*',     —  d^x  -\- m  —  n, 

30.  «  —  3^  +  2;^,     4«  +  7y  —  3;^,     «/  +  ^  —  5fl^. 

Subtraction. 

30.  Def.  Subtraction  consists  in  expressing  the 
difference  between  two  algebraic  quantities. 

Example  1.  Let  us  have  to  subtract  5  —  2  from  11. 
Since  5—2  =3,  we  must  subtract  3  from  11,  leaving  8. 
But  this  is  the  same  thing  as  first  subtracting  5  and  adding 
2.     Hence  the  result  is 

11-5+2, 
and  the  sign  of  2  is  changed  from  —  to  -\-, 

Ex.  2.     Take  a  —  b  from  y. 

It  is  plain  that  if  we  subtract  a  from  y  we  shall  subtract 
b  too  much,  because  a  —  b  i^  less  than  a.  But  if  after  sub- 
tracting a  we  add  b,  we  shall  make  the  answer  right.  Hence 
the  answer  is 

y  -a-^b. 

In  this  answer  the  positive  symbol  a  has  the  sign  —  in 
the  remainder  and  the  negative  symbol  b  the  sign  +. 

We  thus  derive  the  following  rule  for  subtraction: 

31.  Rule.  Change  the  signs  of  all  the  terms  to  be  sub- 
tracted, or  imagine  them  to  be  changed,  and  then  proceed  as  in 
addition, 

Numerical  Examples. 
The  learner  should  first  practice  on  the  purely  numerical  exercises 
until  he  sees  how  the  rule  leads  to  correct  results.    Each  result  should 
be  proved  by  showing  that  the  answer  is  right. 

From  21  +  10 
Take     9-5 

Rem.   12  +  15. 


24  ALOEBBAIG  OPERATIONS. 

Operation.  Imagine  9  to  have  the  sign  minus.  We 
must  by  §  29  subtract  it  from  21,  leaving  12.  Imagining 
5  to  have  the  sign  +,  we  must  add  it  to  10,  making  +  15. 
So  the  answer  is  12  +  15. 

Proof.  Minuend       =  21  +  10  =  31 

Subtrahend  =    9  —    5  =    4 


Remainder  =  12  +  15  = 
which  is  right. 

From    10  +  6           10  +  6           10  + 
Take      9                    9-2            9  - 

37, 

6 
4 

10+6 
9-    6 

Rem.      1  +  6            1  + 
From    25-3 
Take    16  +  1 

8            1  +  10          1  +  12 

27-5            29-7 
15  +  2            14  +  3 

Rem. 

9-4 

12-7 

31-9 
13  +  4 

33-4 

12  +  5 

35-  13 

11  +  6 

Algebraic  Examples. 

From       3a;  —  ^ay  +  55  +  c 
Subtract   x  —  lay  +  85  +  ^. 
We  write  the  minuend,  Zx  —  ^ay  +  55  +  c 

and  subtrahend,  with  signs  changed,  —  x  -\-  lay  —  Sb  —  d 


Result,  applying  Rule  §  29,  2x  +  Say  —  3b  -\-  c  —  d 

From  a  -\-b  subtract  b  +  y. 
«  +  5 

a-y 

EXERCISES. 

1.  From  Ix  —    ^bxy  —  12cy  +    85  +  Sac 

Take  2x  +    75^:^  -    Scy  -    5b  -  2d 


Difference,  6x  —  llbxy  —    4:cy  +  135  +  Sac  +  2d 
Note.     After   the  beginner  is  able  to  operate  by  simply  imagining 
the  signs  changed,  he  need  not  actually  change  them. 

2.  From    8«  +  95  -  12c  -  18c?  —  45a;  +  Sexy 
Take  19a  -  75  -    86-  -  25^^  +  3a;    -  4?/ 


SUBTBAGTION.  25 

3.  From  267z  +  201;^'  +  d2y  +  S6ax  -    6 
Take    UOz  —    S2z'' -\- 202/ -j- 92ax -{- U 

4.  From  8^^  +  lU  subtract  6a  +  10^. 

5.  From  7a  —  3b  —  c  subtract  2a  —  3b  —  3c. 

6.  From  8^^  —  2^  +  3c  subtract  4:a  —  6b  —  c  —  2d. 

7.  From  2x''  -8^-1  subtract  5x'  -  6x -{- 3. 

8.  From  4a;*  -  3x'  -  2x'  -  7x  -{-  9    subtract    x*  -  2x' 
-  2x'  +  7a;  -  9. 

9.  From  2a;''  —  2ax  -\-  3a^  subtract  x^  —  ax -\-  a^. 

10.  From  x""  —  3xy  —  y^  -\-  yz  —  2z^  subtract  x"  +  2xy 
+  hxz  -  3^'  -  2z\ 

11.  From  5a;'  +  6xy  —  12xz  —  ^y''  —  lyz  —  6z'  subtract 
20;=^  -  'Txy  +  4a;;2  -  3y'  +  6yz  -  6z\ 

12.  From  a'  -  3a'b  +  3ab''  -  b'  subtract  -  3ab''  +  b\ 

13.  From  7a;'--  2a;''+  2a;  +  2  subtract  4a;'-  2a;'-  2a;  -  14. 

14.  From  6a'^b—6mn-\-  19xy  subtract  Ha^b-^-Smn-^-lOxy—c. 

15.  From  8{a-\-b)-i-12{m-^n)  subtract  3(a+^)+5(m+70- 

16.  From  9{m^n)  —  6{p  -f-  q)  subtract  5(m  +  n)  —  8(p-\-q) 
+  3  (a;  +  y).  Ans.  4(m  +  ^)  +  2{p  +  ^)  —  3(a;  +  y). 

17.  From  9a  +  12y  +  16{a  +  y)  subtract  12(a  -f  y)  —  6y 
f  10^. 

18.  From    19^   +  23^  _  18-  +  ^  subtract  19^  -  23^ 

b  d  y  b  d 

y 

19.  From  a-\-b  subtract  a  —  b. 

20.  From  x  +  2y  subtract  x  —.  2y. 

21.  From  2a;  —  2y  subtract  x  -\-2y. 

22.  From  a  -\- b  subtract  b  -{-  a. 

23.  From  2jo  +  a;  subtract  x  -\-  2p. 

24.  From  7n  -{- n  -{-  x  subtract  m  —  n  -\-  x. 

25.  From  a  —  b  subtract  b  —  a. 

26.  From  a -\- b  —  c  take  a  —  b  -\-  c. 

27.  From  a  —  b  -\-  c  take  a-\-b  —  c. 

28.  From  1  -m-^-ni'  take  1 -\- m -\- m\ 

29.  From  a'  -  ab -\-  V  take  b""  -  3ab  +  a\ 

30.  From  2'  +  2;  +  m^i  take  z  —  z^  -\-  nm. 

31.  From  ab  take  Ja;  from  4  .  7  take  7  .  4. 


26  ALGEBBAIG  OPERATIONS, 


Clearing  of  Parentheses. 

33.  Plus  Sign  before  Parentheses,  If  a  poly- 
nomial is  enclosed  between  parentheses  and  preceded 
by  the  sign  +,  the  parentheses  may  be  removed  and 
all  the  terms  added  without  change. 

Example.  22  +  (8  —  15  +  12)  is  the  same  as  22  +  8  — 
15  +  12. 

Proof.       22  +  (8  -  15  +  12)  =  22  +    5  =  27 
and  22  +    8-15  +  12   =  42  -  15  =  27. 

EXERCISES. 

Clear  of  parentheses  and  add: 

1.  5^  -  3^  +  2c  +  (-  2a  +  9^  -  3c)  +  {la  -  lU). 

2.  36m  +i?  -  g  +  (24m  -  5^)  +  (-  42m  +  Sp). 

3.  a-^l)-\-{a+b-c)-\-{a-i-{-c). 

4.  a-\-'b-c-\-{a-l)-Yc)^{-a^h-\-c), 

5.  a-h-^{l)-c)-]-{c-a). 

6.  2:^  -  4^/  +  2^  +  (-  ^x  +  2?/  +  %z). 

7.  m  +  ^  -  2j(?  +  (m  -  2w  -^p)  +  {-  2m  +  ^  -\-p). 
m       a 
n       b 

9.  m-\-^n-{-  {7n  -  2n)  +  {p -{-  2q)  +  (p  -  2q), 
10.  x-{-y-z-^(y-\-z-x)-\-(z-^x-y). 


-fH-14-3' 


m       ^    I    f  ^  _  ^  ]    I    ( ^  _  !!^ 
'  ~n      b       \b      yJ\y      ti 

12.  ax-by-\-  (2by  -  2ax)  +  {3by  -  3fl^a;). 

13.  277171  —  6xy  -\-  {577171  —  2xy). 

33.  Minus  Sign  before  Parentheses.  If  the  paren- 
theses are  preceded  by  the  sign  —  we  remove  them  and 
change  the  sign  of  each  enclosed  term  (§  31). 

Note-  Remember  that  if  a  term  has  no  sign  before  it,  then  it  is 
positive,  and  must  by  this  rule  be  changed  to  negative. 

EXERCISES. 

1.  a -3b  —  {2b  -  5a)  -\- {m -\- Sa  -  b)  —  {a  -  m). 
Solution.     Changing  the  signs  by  the  rule,  where  the  parentheses 
are  preceded  by  the  sign  — ,  the  result  is 


TO  MULTIPLY  A  MONOMIAL  BY  A  NUMBER  27 

a  —  db--2b-\-5a-{-m-\-da~b  —  a-{-7n. 
The  coefficients  of  a  are  +  1  +  5  +  3  -  1    =    +8. 
"    6   "    --  3  -  2  -  1  =    -  6. 

«  "  "  m   "  +l-]-l  =    +2. 

Therefore    Ans.  =  8a  -  6^*  +  2m. 

2.  x-3y  -  (27/  -  6x)  -}- {z -]- 3x  -  y)  ~  (x  -  z). 

3.  Qm  +  9/i  —  (m  —  h)  +  [h  —  m)  —  (2m  +  2h), 

4.  a  -  (-  «  +  ^)  -  (-  fl^  +  c)  -  (-  ^  +  ^). 

5.  X  —  y  —  {y  —  x)  —  X  -{-  ^y- 

6.  «  +  Z>  -  26?  -  (c  -  2^>  +  fl^)  -  (^  -  2fl^  +  c). 

7.  2z  —  X  —  y  — \z  —  2y  ^-  x)  —  {y  —  2x  -{-  z). 

8.  «a;  —  (dax  -\-  2by)  +  9<^^- 

9.  m  —  n  —  {a  —  h)  —  {c -[-  d)  —  {e -\-  g), 

10.  2am  +  {3am  —  ?i)  —  (n  —  3am)  +  2n. 

11.  77i  +  "Ih  -  (7A  -  7/1')  +  {^k  -  ^^0- 

12.  13px  —  qy  —  {~  qy  —  rz)  —  {rz  +  2px). 

13.  2«jo  -  3bq  —  {Sap  +  SJ^^)  +  (Qap  -  c  -  d). 

14.  fl^  —  (:r  +  ?/  —  ;2;  +  ^^  —  ^  +  ^)* 

15.  i?;  —  (a;  —  ?/  +  ?^  —  w). 


Sectiot^  III.    Mtjltiplicatiois". 

To  Multiply  a  Monomial  l>y  a  Number. 

34.  EuLE.  Multiply  the  coefficient ^  and  to  the  product  affix 
the  algebraic  symbols  of  the  multiplicand. 

Example  1.     Multiply  3ax  by  7.  3ax 

7 


Ans.  21ax 
Ex.  2.     Multiply  ab  by  5. 
Here  the  coefficient  of  ab  is  1  and  the  product  is  ^ab, 

EXERCISES. 

Multiply: 

1.   "laby  by  8.  2.   hmx  by  6.         3.  ^pqr  by  7. 

4.  22J(«  +  ^)  by  2.  Ans.  44J(a  +  x). 

5.  3(6'  +  .v)m  by  9.  6.  2(«  +  ^)  by  3. 

7.  a  +  J  by  7.  Ans.   7(«  +  b). 

8,  iz;  —  ^  by  8.  9.  2(5  -  c)  by  12. 


28 


ALQEBKAIG  OPERATIONS. 


To  Multiply  one  Monomial  by  Another. 

35.  Rule.     Multiply  the  coefficients,  and  affix  to  the  pro- 
duct  all  the  symbols  loth  of  the  multiplier  and  7nultiplicand, 
Example.     Multiply  12amx  by  lacy. 

Multiplicand,  12amx 

Multiplier,  Hacy 

Product,  S4:aacmxy  =  S4:a^cmxy 

Solution.  7  X  12  =  84  is  the  product  of  the  coeflQcients, 
to  which  we  affix  the  symbols.  The  symbol  a  being  taken 
twice  as  a  factor,  we  write  a\ 


Multiply: 

1.  dab  by  12mx. 

2.  5amy  by  11^5. 

3.  '7ahj  Qamx. 

4.  6{m  -\-  n)  by  3w. 

6.  %(x-{-y)hj<da{x-\-y). 

6.  b{a-l)  by  3c. 

7.  l{x  -  z)  by  ^x  +  z). 

8.  19(m  —  n)  by  h{p  —  q). 
10.  U{c  -  d)  by  9g(r  -  s). 
12.  5m(m—p)  by  6m{m—p). 
14.  ab  by  ba. 

16.  2(a  +  b)  hj2{a-b). 
18.  mn{x-\-y)  by  mn{x—y). 


EXERCISES. 


Ans.  65a^bmy, 

Ans.  18m{m  -\-  n). 
Ans.   'l!2a(x  -\-  yY, 

Ans.  21  (a;  +  2;)  (  a;  —  2;). 

9.  1a{x—y)  by  6(m— :^). 

11.  24^-^)  by  db(p-q). 

13.  76?(^>  —  c)  by  a. 

15.  a;?/  by  3(a;  +  y). 

17.  5m/zjo  by  4:mnq. 

19.  mp{a-\-b)  by  ^^'(a+J). 


20.   6fl^^(a;  +  1/)  by  7«c(a;  +  y). 

36.   Z7se  0/  Exponents.     Let  us  have  to  multiply 


By  definition. 
Therefore 
Therefore 


a'  =  aaa; 
a^  =  aa. 


a^  y  a^  =  aaa  X  ««  =  aaaaa  =  a' 


MULTIPLICATION  OF  MONOMIALS.  29 

Hence 

EuLE.     The  exponents  of  like  symbols  in  the  factors  must 
he  added  to  form  the  exponent  in  the  product. 

Note.    In  the  following  exercises  the  pupil  may  advantageously 
go  through  the  above  process  in  each  case,  until  he  fully  understands 


reason  for  it. 

EXERCISES. 

Multiply: 

1.  x'y.x\ 

^.  (^+^rx(^+2/r. 

Atih. 

(x\ 

3.  a'  X  a\ 

4.  V  X  l\ 

6.   {a  +  hY  X  («  +  l)\ 

6.  (m  +  nY  X  (m  +  n)\ 

7.  2A*  X  3/i\ 

Ans. 

Qh\ 

8.  3m  X  6m' 

Ans. 

18m*. 

Where  there  is  no  exponent,  the  exponent  1  must  be  understood. 

9.  Wh'  X  %a'h.  Ans.  lQa'b\ 

10.  brn^n  X  5mw^ 

11.  ^ahx  X  ba'lfxy, 

12.  Sa'^^V  X  ^a'hxy. 

13.  86j'(m  +  7i)  X  9a'(m  +  t^)'.         Ans.  72«X^  +  ^)'« 

14.  2(2;  +  ?/)  X  3(.c  +  ^)'^«^. 

15.  3(^  +  ^r2:/x3(j9  +  ^)^^. 

16.  6(A  +  ^)%»?^  X  {h  +  Jcym7i'x, 

17.  2^c(^  +  c)  X  ^V'(^  +  c)^. 

18.  M'x(d  +  a;)''  X  2^2;'(^  +  x)\ 

19.  3(m  +  ^)  (i?  +  qY  X  2(m  +  ^^)'  (p  +  ^). 

20.  4(«  +  bY  iff  +  ^)'  X  6{a  +  J)*  (^  +  h). 

37.  C'flj^e  0/  more  than  Tzvo  Factors.  When  there  are 
three  or  more  factors,  multiply  two  of  them,  then  multiply 
that  product  by  the  third,  etc. ,  until  all  are  multiplied. 

All  the  preceding  rules  may  be  combined  in  the  following 

Rule  for  the  Multiplicatiok  of  Moj^omials.     Fonti 

the  product  of  the  numerical  coefficients  and  to  it  affix  all  the 

symbolic  factors,  giving  each  the  sum  of  its  exponents  in  the 

several  factors. 


30  ALGEBRAIC  OPERATIONS. 


EXERCISES. 

Multiply: 

1.  ab  X  ic  X  ca. 

Solution,     a,  h,  c  are  the  factors,  and  the  sum  of  the  ex- 
ponents of  each  is  2.     Hence 

ah  X  he  X  ca  =  o^V&.  Ans. 

2.  xy  X  yz  X  zx. 

3.  mn  X  np  X  pm. 

4.  ah  X  ac  X  ad.  Ans.  a^hcd. 

5.  2mx  X  Smy  X  mz.  Ans.  Qm^xyz, 

6.  2mx  X  dmy  X  Qmz. 

7.  a  X  ah  X  ahx  X  ahxy.  Ans.  a^h^x'y. 

8.  2«  X  ^ab  X  4:ahx  X  hahxy. 

9.  "^a  X  {m -\- n)  X  h.  Ans.  %ah{m  +  w). 
Kemark.     The  pupil  will  treat  quantities  in  parentheses  as  mo 

nomials. 

10.  2m  X  {x-[-y)  X  a. 

11.  '2mxlx-\-y)  X  3a. 

12.  ^aX  {a-\-h)  X  6ax. 

13.  46j'm'  X  (fl^  +  ^)  X  2amY         Ans.  8«'m*(a  +  h)y. 

14.  Stt''^'  X  h'c'  X  c'a\ 

15.  mVi"  X  ny  X  p'm\ 

16.  a"  X  a^  X  a, 

17.  a^  X  a'^'a;  X  a\  Ans.  a'^'^r. 

18.  am  X  a'n  X  a'p,  19.  h'd  X  d'c  X  c'h. 
20.  h^  X  2^'  X  U\  21.  a^'^  X  "lax^  X  2b'x\ 
22.  m  X  nf  Xtri"  X  m\                 23.  wa;  X  wa;'  x  mx^, 
24.  2«a;  X  2ay  X  2az  X  2xyz.       25.  2a'x'  X  Say. 

To  Multiply  a  Polynomial  by  a  Monomial. 

38.  Rule.     Multiply  each  term  of  the  polynomial  hy  the 
monomial  and  take  the  algehraic  su?n  of  the  products. 
Numerical  Example.     Multiply  3  +  4  +  5  by  3. 
3+4+5     =     12 
3  3 

9  +  12  +  15     =    "36 
We  see  by  this  example  that  the  product  of  the  sum  of 
several  numbers,  as  3,  4  and  5,  is  equal  to  the  sum  of  the  pro- 


TO  MULTIPLY  A  POLYNOMIAL  BY  A  MONOMIAL.     31 

ducts  found  by  multiplying  each  number  separately,  because 
we  get  the  same  result,  36,  in  either  case. 

Algebraic  Example.    Multiply  2a  -\-  dixy  +  ^^^  ^J  '^am. 
Multiplicand,    2a      +  Sbxy        +  ^^(^ 
Multiplier,  7am 

Product,  14a^m  -|-  ^lahmxy  +  Boa^mc, 

Operation.     Each  term  is  multipHed  separately,  by  the  last  rule,  and 
the  sum  taken. 

EXERCISES. 

Multiply: 

1,  a  -\-  b  hj  x\  2,  m-\-  n  by  a, 

3.  a-{-chj2x,  4.  2a  +  3c  by  32;. 

5.  a -\- 2b -{-de  by  2ai/,  6.  2a-{-3b-{-  4:C  by  abc. 

7.  Sab  +  4«c  by  babe.  8.  4:ax  +  5by  +  7cz  by  Sxyz, 

9.  «m  +  Z>;?  +  eg  by  2abc,  10.  6fl!m  +  8^/^  +  9c/^  by  7«a;. 

11.  6x-{-6xy-\-'7xyz  by  «a:?/;2;.  12.  2a:  +  ^2/  +  ^^  +  ^^  ^J  ^^^• 

13.  m-{-7nn-\-mnp  by  m?z^.  14.  ^  +  2g  +  3r  by  2apq. 

15.  ProYB  the  equations 

9(8  +  7)  =908  +  9.7. 

6(5-3)  =  6  o  5  -  6  .  3. 

2(1  +  1)  =  2  +  2. 
2(1  +  1  +  1)  =  2  +  2  +  2. 

7(9-5)  =7o9-7.5. 

6(6  -  3)  =  6'  -  6  ,  3. 

q\q  -5)  =6. 

39.  If  any  terms  of  the  polynomial  are  negative 
their  product  by  a  positive  multiplier  is  negative. 

EXERCISES. 

Multiply; 

1.  a  —  bhj  m,  Ans.  ma  —  nib, 

2.  m  —  n  by  a. 

3.  a  —  nhj  b. 

4.  2«  —  Sn  by  c. 

5.  —2a-\-  3n  by  2J.        Ans.   —  4a5  +  6Jw. 

6.  a  —  b-{-c  —  dhyx. 

7.  a  —  2Z>  +  3c  -  4c?  by  ab. 

8.  2a  -  3^  +  46-  by  abc. 

9.  —  a-\-  b  —  dc  by  aa;?/. 


82  ALGEBRAIC  OPERATIONS. 

10.  '7ab  -\-  2bx  —  ax  by  2abx, 

11.  —  6am  -\-  7bn  —  Sep  by  mnp, 

12.  —  a^x  +  b'^y  -\-  &z  by  xyz. 

13.  %rrei  -  Zn^j  +  4j9'X;  by  ijK 

14.  7«''m  —  %lfn  —  l^c^jt?  by  ^mnp, 

15.  12^2;''  -  10Z>^=^  -  Icz"  by  8a5c. 

16.  -  8«m"  +  9^?^'  +  lOc/  by  2abc. 

17.  3«J  +  2^c  —  cfl^  by  a^c. 

18.  ^a'b  -  Ub'  by  12a;2^. 

19.  —  l^ab  —  Sbc  -  9ca  by  '7abc. 

20.  —  xy  —  2yz  —  32;a;  by  5xyz, 

21.  «^'*  —  ^'^o;  by  axy.  Ans.  a'*a:^y  —  a^x*y. 

22.  -  2mb -i- 2m'b' hj  3mbr. 

23.  —  3«'m  —  2Z>'?^  by  3a2»mw. 

24.  Sfl^rc  —  36?'^?/  ^y  ^f^^y* 

40.  Indicated  Multiplications.  Tlie  multiplica- 
tion of  a  polynomial  may  be  indicated  by  enclosing 
it  in  parentheses  and  affixing  or  prefixing  tlie  multi- 
plier. 

EXERCISES. 

Execute  the  following  indicated  multiplications: 

1.  3(2fl^  —  hbx  —  c).  Ans.   6a  —  lobx  —  3c. 

2.  2(a—2b  +  3cy). 

3.  4:(mx  —  2ny  +  3pz). 

4.  4:a{ab  -\-  ac  —  ad).  Ans.  4:a^b  -\-  ^a^c  —  ^a^d, 

5.  2m{m^  +  '^'^^  +  ^*)' 

6.  3Z''^(Z''^^''  -  cY).  Ans.  3Z>*a;'^  -  3^>Vy. 

7.  c^c'x''  -  c?/''). 

8.  cy(c'  +  2/'). 

9.  2«"^^(3aa;  -  2%). 

10.  2a(«  +  ^)a2'.    '  Ans.  2a'b -\- 20^^. 

We  first  multiply  the  two  factors  2a  and  db  which  are  without  the 
parentheses,  making  2a'^b.  Then  we  multiply  each  term  within  the 
parentheses  by  this  product. 

11.  2m{m  +  n)mn.  12.  3h{h  —  hyik. 
13.  2m{m  -  n)mn.  14.  4:a\a  -  ¥)b\ 
15.  2^'^(2:  -  i/)a:.v.  16.   Vi\2m  -  3n)k. 
17.  Sb{b'  +  A)^^  18.  3a'^^^(2a  -  3h)h. 
19.  2mn\m.  +  27^)3m.  20.  2a^(«''  —  ^'')2rt'^. 


TO  MULTIPLY  A  POLYNOMIAL  BY  A  MONOMIAL.       33 

41.  Negative  Multipliers. 

Rule.      When  the  7nuUipUer  is  negative,  change  the  signs 
of  all  the  terms  of  the  product. 

The  reason  for  this  rule  will  be  given  in  Course  II. 

EXERCISES. 

1.  _  3(^  _  2§  +  3c  -  4:d).  Ans.  -  3fl^  +  6J  -  9c  +  12d, 

2.  —  2{m  —  2n  +  3p  —  4:q). 

3.  -a{x-{-y-z). 

4.  —  a^{ax  —  hy  -{-  cz). 

5.  —2al){—^ax  +  21)y).  Ans.  Mix  —  ^a¥y, 

6.  -  %al{a^  -  V)a.  Ans.   -  ^a'l  +  %a^h\ 

7.  -  2mn{7)i^  -  n')n,  8.   -  3mx{x'  -  y')xy, 

9.   —  8p{x  —  y)xy.  10.   —  2a(ax  —  a'y  +  a^z). 

11.   ^rn{x-y-z)m\  12.    -  U'h\h  -  2h')h. 

13.   -  4/m^(«  -  7i  -  2m)2am.       14.   -  4a^^>V(a^  -  ^'^  +  c''). 

43.   Comlination  of  Multiplication  and  Addition, 

EXERCISES. 

Execute   the    following    indicated    multiplications,    and 
simplify  the  results  by  addition: 

1.  2a{^x  —  4.y)  -  a{x  +  2y)  —  3«(—  x  —  Sy), 
Work:         2a{'Sx  —  Ay)  =    Qax  —  Say 

—  a{x    -\-2y)  =  —  ax  —  2ay 
—da{—x—3y)  =    Sax  +  9ay 

Sum  =    Sax  —  ay        Ans. 

2.  3(fl^  +  b)-  2{h  +  c)  -  (c  -  a). 

3.  2«(o^  +  ^)  +  2^(^  +  a).  Ans.  2«''  +  ^ah  -\-  2h\ 

4.  37i(^  +  m)  +  3m(7/i  +  h). 
6.  2«(a  +  ^)  -  2J(^'  +  a). 

6.  -  ^72(a  -^l-c)  -  2a{m  -  V). 

7.  a{l)  -c)  -  l(c  -a)  -  c(a  +  i), 

8.  a:(,?/  -  2;)  +  y{z  -  x)  +  ;^(2;  -  y). 

9.  m'^(a''  -  Z*)  -  a'ifn'  -  b)  -  2h{m'  +  «'). 

10.  -  3(3rc  -  2y)  +  2x  -  5?/  +  4(8a:  -  2?/). 

11.  2a;(a:''  +  2x)  -  Sx^x  -  2)  -  4(6  -  x'), 

12.  2a''(aj''  -  a')  -  2x\a''  -  x'). 

13.  2x{x  ~y)-j-  2y{x  -  y). 


34  ALQBBnjLlG  OPERATIONS. 

14.  ^h'  +  U{Vi  -  7)  -  U{U'  -n-  2). 

15.  4?i'  —  bn{n  —  3)  —  n{n  —  2). 

16.  2x{x  —  «/)?/  4-  ^^(y  —  ^)^' 

17.  2a;(a  -  Z>)  -  2a{x  -  h) -\-  %bx. 


Section  IY.     Divisiot^". 

43.  Def.  The  quotient  is  that  quantity  which 
multiplied  by  the  divisor  will  produce  the  dividend. 

The  dividend  is  said  to  be  exactly  divisible  by 
the  divisor  when  it  contains  the  divisor  as  a  factor. 

44.  First  Principle  of  Divisiojs".  A  product  may  he 
divided  hy  any  of  its  factors  by  simply  removing  them. 

Example  1.  3.4,  which  is  12,  may  be  divided  by  4  by 
removing  the  4  leaving  3  as  the  quotient. 

Ex.  2.  The  product  abc  may  be  divided  by  h  by  remov- 
ing b  leaving  ac  as  a  quotient. 

Remark.  When  all  the  factors  of  the  dividend  are  re- 
moved the  quotient  is  not  0  but  1. 

Ex.  3.  2.3,  which  is  6,  divided  by  2  .  3,  which  is  6,  is  1; 
ah  divided  by  ab  is  1. 

Hence,  to  divide  one  expression  by  another,  which  is  an 
exact  divisor  of  it: 

Rule.  Remove  from  the  dividend  those  factors  whose  pro- 
duct forms  the  divisor.  The  product  of  the  remaining  factors 
will  he  the  quotient.  If  all  the  factors  are  removed  the 
quotient  is  1. 

EXERCISES. 

1.  Divide  3  .  4  .  7  by  3  .  4. 

2.  Divide  6  .  8  .  9  by  8.  Ans.  6  .  9  =  54. 

3.  Divide  ab  by  a. 

4.  Divide  ^amn  by  2m.  Ans.  %an. 

5.  Divide  l%nxy  by  4?/. 

6.  Divide  l^bpr  by  ^br. 

7.  Divide  l{a  -\.b)hja^b.  Ans.  7. 

8.  Divide  5m{m  +  w)  by  5m. 

9.  Divide  6p{r  +  s)  by  3(r  +  s). 

10.  Divide  Su{h  +  g)  by  4:u{h  +  g). 

11.  Divide  12d{x  +  y)fhj  4/. 


DIVISION.  35 

12.  Divide  a{x  +  y)  {m  -\-  n)  by  a{m  -\-  n), 

13.  Divide  15{f -{- g)2x{h  +  g)  by  6x. 

14.  Divide  16(r  +  s)4:mn{u  +  ^)  by  8m(?^  +  ^)- 

45.  i^w^e  o/  Exponents,     Let  us  have  to  divide  a""  by  a". 

We  have 

a^  =  aaaaa, 
a^  =  aaa. 

Hence  by  the  preceding  rule  a^  ~  d  z=.  aa^=  a^. 

Here  2,  the  exponent  m  the  quotient,  is  obtained  by  sub- 
tracting the  exponent  of  the  divisor  from  that  of  the  dividend. 
Hence 

EuLE.  Tlie  exponent  of  any  symbol  in  the  divisor  is  to  be 
mbtr  acted  from  the  exponent  of  the  lihe  symbol  in  the  dividend. 

Remark.  Iq  applying  this  rule  remember  that  a  quantity  without 
any  exponent  is  to  be  considered  as  having  the  exponent  1. 

EXERCISES. 

Divide: 
1.  m^  by  m".  Ans.  m^.     2.  m^  by  m, 
3.  m'  by  m^,  4.   (a  +  b^  by  o^  +  5. 

5.  am^  by  am.  6.  a^m^  by  am.     Ans.  am^. 

7.  p^(i'  \iypq\  8.  12/7i'  by  4^.     Ans.  ^(jh\ 

9.  l^g^h  by  5^.  10.  ^km'^n^  by  ^7imk.     Ans.  dmn\ 

11.   8AFm^by2mk       12.   67^X«  +  ^)' ^J  ^^K^  +  ^)- 

13.  'ir\x  +  ^)'  by  t{x  +  ^)V. 

14.  14?^  V(^^  -h  ?;)  by  2^^. 

15.  lb{a  +  Z*)^  (a;  +  vY  by  5(«^  +  2*)  (a;  +  ^). 

16.  lQab''c\x-\-yyhj  ^{x-\-y)abc. 

17.  3a'm'(jy  -  qY  by  da'm'{p  -  qY. 

18.  5pq^{m  —  7i)  hj  pq\m  —  n). 

19.  pa^xy^z  bv  aa:?/;?. 

20.  12(«  -  Z<)'3^^(r  -  sY  by  9(fl^  -  J)  (r  -  s)q. 

21.  9(«  -  Z»)  (r  -  5)  by  9(r  -  s)  (a  -  3). 

22.  x""  by  a:".  Ans.  a;"*-r^ 

23.  fhjy^.  } 

24.  (:r  +  hYr^  hyr^{x  +  A)". 

25.  a'^x'''  by  a^z". 

26.  {m  +  yi)2'^2;^  by  (m  +  nYx"", 

27.  (a- J)^^by(a-Z')"  +  ^ 


36  ALGEBRAIC  OPERATIONS. 

To  Divide  a  Polynomial. 

46.  Rule.  Divide  each  term  of  the  polynomial  separately. 
The  sum  of  the  separate  quotients  will  he  the  quotient  required. 

Example.     Divide  m^a  +  ma^  by  m. 

We  first  divide  m^a,  leaving  the  quotient  ma.  Then  we 
divide  ma^,  leaving  the  quotient  a^ 

Hence  the  answer  is 

m^a  +  ma^  -^  m  =  ma  +  a^, 

EXERCISES. 

1.  iri^x  +  mx^  -^-  m  =  what  ? 

2.  a^y  +  ay"  -^  a. 

3.  9bY  +  12by' ^  3by. 

4.  8p'x -{- 4:p'y  +  16p*z -^4:p\ 

6.  9abx'  +  I'Zab'x  +  ISa'^a;  ^  3aJa;. 

6.  5/i'(m  +  w)  +  10y\m  -j- n)  -^  m -\- n. 

7.  (a;  +  2/)  (:?;  -  2/)  -  (x  -\- yY -^  x -\- y. 

8.  («  +  Z*)  («  -  by  +  («  +  ^)'  («^  -  ^')  4-  (a  4-  ^)  {a  -  h). 

9.  3^;  -  6a;'  +  3^'  -^  'dx. 

10.  6m' (a;  +  y)  -  12m' (a;  —  y)  -^  6m. 

4*7.  /S^^'^7^  o/"  the  Quotient.  When  the  divisor  is 
positive,  the  quotients  of  negative  dividends  must  be 
negative. 

Example.     If  we  have  to  divide  12  —  8,  that  is  4,  by  2, 

the  quotient  from  the  separate  terms  will  be  6  —  4.     It  is 

evident  that  the  true  answer  is  6  —  4,  and  not  6  -f-  4.     Hence 

12 -8^2  =  6-4  =  2. 

This  principle  may  be  expressed  in  the  following  form: 

Like  signs  give  -\-  ;    unlike  signs  give  — . 

EXERCISES. 

1.  9a"x^  —  12ax"  -f-  3ax.         Ans.  Sax'  —  4ic. 

2.  8m'a;  —  4:m^y  —  lQm*z  -^  4m'. 

3.  9pqx"  —  12pq"x  +  15pq^x  ~-  Spqx. 

4.  r"{m  —  nY  —  5r^(m  —  w)'  -=-  r{m  —  ny, 

5.  V(g  -  hy  -  2b{ff  -  ny  -=-  l{g  -  li)\ 

6.  bx  +  cx^  —  dx^  —  gx*  +  hx^  -^  x. 

7.  hax  —  lOa'^y  +  ba^z  -^  6a. 

8.  dmu  —  Qm'^w  —  9mw"  -r-  3m. 


FACTORS  AND  MULTIPLES.  37 

9.  5m' (ic  —  y)  —  l^m{x  —  yY  -^  bm{x  —  y), 

10.  2(m  +  x)y  -  Qy\m  +  xf  -^  %y. 

11.  Zx\x  +  A)  —  9^;^^  +  lif  -^  3(ic  +  }i)x. 

Factors  and  Multiples. 

48.  Def.  A  prime  expression  is  one  wMch  has  no 
factors  except  itself  and  unity. 

Def,  A  composite  expression  is  one  wMch  can  be 
expressed  as  a  product  of  two  or  more  factors. 

Examples.     The  number    7   is  prime  because  no  two 
numbers  multiphed  together  will  make  7. 

21  is  composite  because  it  is  equal  to  7  X  3. 

a-\-x\%  prime. 

a"  +  ax  is  composite  because  it  is  equal  to 
(a-\-x)y^a 
which  is  a  product. 

49.  Bef.  The  degree  of  a  monomial  is  the  number 
of  literal  factors  which  it  contains. 

Example.      The  monomial  Zax^  is  of  the  fourth  degree 
because  it  contains  the  literal  factors  a,  x,  x,  x, 

EXERCISES. 

"What  is  the  degree  of  these  expressions? 

1.  hx.  2.  hx\  3.  ^'bx\ 

4.  ^c{a-^x).         5.   1c{a^x)\        6.  mV. 

50.  Def.  To  factor  an  algebraic  expression  means 
to  express  it  as  a  product  of  several  factors. 

Problem.  To  factor  a  number  or  algebraic  expression. 
EuLE.  Find  hy  trial  what  prime  number  or  expression 
tvill  divide  it ;  then  find  what  number  or  prime  expression 
will  divide  the  quotient.  Continue  the  process  until  a  prime 
quotient  is  reached.  The  divisors  and  last  quotient  will  be  the 
factors  required. 

Example.     Factor  the  number  420. 

420  -^  2  =  210; 

210  ^  2  =  105; 

105  -^  3  =    35; 

35  -^  5  =      7. 

Hence 

420  =  2\  3 .  5 .  7,  which  are  the  factors  required. 


38  ALQEBEAIO  OPERATIONS. 

EXERCISES. 

Express   the   following    numbers   as   products   of    prime 
factors: 

1.     36.     Ans.  2' .  3^  2.     12. 

3.     28.  4.     81.     Ans.  3*. 

6.  256.  6.  324. 

7.  72.  8.  140. 
9.     56.                                10.     82. 

11.  100.  12.     52. 

51.  Factoring  Algebraic  Expressions, 
Example  1.     To  factor  ¥  -\-  hy. 

We  see  that  h  will  divide  both  terms.     Dividing  by  1)  the 
quotient  is  ^  -|-  ^  which  is  prime;  therefore 
V  -^hy  =  h{h  +  y), 
Ex.  2.     «'  -  a'y  =  a\a  -  y). 

EXERCISES. 

Factor: 

1.  ni^  -\-  mn. 

2.  lx-\-hy, 

3.  ix-^ly  -\-  hz.  Ans.  l{x -\-  y  -\-  z). 

4.  ex  —  2cy  -\-  Scz.  Ans.  c{x  —  2y  -{-  3z). 

5.  hp  —  dhg  -^  bhr. 

6.  hp  —  ahg  +  hhr. 

7.  2/i>  -  2a/iV  +  4:Z>AV.  8.  n-^n"  -\-  7i\ 

9.  n-  dn''  +  5n\  10.  Ux'  -  8i'x  +  12^>a;. 

11.  Shy  +  6/^V  +  12^Xv^-  1^-  4^'  -  ^«^'  -  l^^'- 

13.  4:mn'  +  8m' w'  +  ^m'n. 

14.  («  +  ^•)a;  -\-{a-\-  b)y.     Ans.  (a  +  Z>)  (a:  +  y), 

15.  (w  +  w)a;  +  {m  +  w)y.  16.   {g  +  >^)?^  -  (^  +  ^)v- 
17.  a(x  -y)-{-  h(x  -  y),  18.  a\x  -^  y)  -  h{x  +  y). 


Highest  Common  Divisor. 

53.  Def.  A  common  divisor  of  several  quantities 
is  any  quantity  which  will  divide  them  all  without  a 
remainder. 

Def.  The  highest  common  divisor  of  several  quan- 
tities is  their  common  divisor  of  highest  degree. 


LOWEST  COMMON  MULTIPLE.  89 

Def.  When  two  or  more  quantities  liave  no  com- 
mon divisor  but  unity,  they  are  said  to  be  prime  to 
each  other. 

Notation,     The  highest  common  divisor  is  written, 
for  shortness,  H.  CD. 

53,  Problem.     To  find  the  H.O.D.  of  several  quantities. 
Rule.     Factor  each  of  the  quantities.      The  continued 

product  of  the  factors  common  to  all,  each  with  its  lowest  expo- 
7ienty  is  their  H.  0.  D. 

Example.    Find  the  H.  C.  D.  of  24,  36  and  48. 
24  =  2^  3;         36  =  2».  3=";         48  =  2*.  3. 

The  common  prime  factors  are  2  and  3;  the  highest  ex- 
ponent of  2  is  2;  .  •.  2^  3  =  12  is  the  H.  0.  D. 

EXERCISES. 

Find  the  H.  0.  D.  of  the  following: 
1.  54;  90;  144.  2.  14;  56;  63;  84. 

3.  72;  108;  132. 

4.  ax\  bx;  dcx\  Ans.  x, 

5.  bni;  hn\  Ih.  6.  2mn^x;  6mn^x*, 

7.  Sx'yz;  dxYz;  15a;>.  Ans.  3x'y, 

8.  VZm'n*;  20mn'',  2^m'n\ 

9.  I){x  —  h);  c{x  —  A).  Ans.  x  —  h. 

10.  m''{m-\-  n)',  mn{m -{- n), 

11.  3m{g-h);  Qm\  12.   6m{g  ^  h);  12{g  -  h). 

Lowest  Common  Multiple. 

54.  Def.  The  multiples  of  any  quantity  are  all 
quantities  which  contain  it  as  a  factor. 

Example.  The  multiples  of  3  are  3,  6,  9,  12,  etc.,  and 
3a,  6i,  15c,  etc.,  and  in  general  all  quantities  which  contain 
3  as  a  factor. 

Def.  A  common  multiple  of  several  quantities  is  any 
expression  which  contains  all  the  quantities  as  factors. 

Example  1.  24  is  a  common  multiple  of  3,  4,  6,  8,  be- 
cause 24  contains  3  as  a  factor,  4  as  a  factor,  6  as  a  factor 
and  8  as  a  factor. 

Ex.  2.  ab^{x  4-  ^)  is  a  common  multiple  of  a,  i,  db,  ai*, 
X  -\-  y,  etc.,  because  it  contains  each  of  them  as  a  factor. 


40  ALGEBRAIC  OPERATIONS. 

Def.  The  lowest  common  multiple  of  several 
quantities  is  their  common  multiple  of  lowest  degree. 
The  lowest  common  multiple  is  written,  for  shortness, 
L.  C.  M. 

^^.  Problem.     To  find  the  lowest  common  multiple. 

Rule.  Factor  each  of  the  quantities.  The  product  of 
the  different  factors,  each  aff'ected  with  the  highest  exponent  it 
has  in  any  one  of  the  quantities,  is  the  L.  C.  M. 

Example.     Find  the  L.  0.  M.  of  "^xif,  3yz  and  6^'^. 

The  different  factors  are  2,  3,  x,  y,  z. 

The  highest  exponents,  1,  1,  2,  2,  1. 

Lowest  common  multiple,  2.3.  x'^y'^z  =  Qx'^y^z, 

Quotients,  dxz,  2x''y,  y^. 

EXERCISES. 

Pind  the  L.  C.  M.  of  the  following  expressions  and  the 
quotients  formed  by  dividing  the  multiple  by  the  several 
quantities: 

1.  4:a'h'c;  ISa'h'c;  Sadc\  2.  2mn^;  3np;  6m>. 

3.  qr;  rs;  st;  qt.  4.  4m?^;  8mj9;  4,np, 

5.  Qgh""',  ^''h\  ^g'h^',  ZghK 

6.  mw':  nr"^',  rs;  srri^. 

7.  2u^vw;  4:uv^tu;  6u^v^w;  6u. 

8.  3Z'(m  +  w);  6h' (m -{- n) ;  3(m  +  ^)'. 

Ans.  L.  0.  M.    =  6b'{m  +  n)'. 

Quotients  :=  21f{m -\-  n),  m  -{-  n  and  25'. 

9.  c^{m-\-n);  c^{m-^ny;  c{'m -{- n)'. 

10.  pq\g  -  h);  p'q{g-h);  p\\g  -  h). 

11.  x\  xy,  xyz',  xyz{x -\- y\ 

12.  12A;  4^;  3/^U\ 

13.  {a-^-l)  {a-iy,  {a-^l)\ 

14.  \a  -  xy-,  U\a  -  1);  W{a  -  1). 

15.  %x  -  yY;  l(x  +  y)  {x  -  y)',  14.{x  +  y)\ 

16.  24(«  -  hyx;  36(«  -  l))x\ 

17.  m;  ni^n;  m^n'^p;  'm*n^p^{m  -\-p). 

18.  ax;  2d^x;  3a^x;  4:a*x. 

19.  a{x  —  y);   a{x -]- y);   a{x  —  y)  {x -{- y). 

20.  abc{2J  —  q);   abed;   ah{p  —  q). 

21.  (m-^n)  (p  —  q);    (p  —  q){m  —  ?i);    {m  —  n)  {m-\- n). 


MISCELLANEOUS  EXERCISES.  41 

MISCELLANEOUS     EXERCISES. 

1.  In  the  division  of  an  estate  A  got  y  dollars,  B  got  c 
dollars  more  than  A,  and  C  got  3  times  as  much  as  A  and  B 
together.     How  much  did  they  all  get? 

2.  A  pedestrian  on  the  first  day  walked  h  hours  at  the  rate 
of  q:  miles  an  hour;  the  second  day  he  walked  the  same  length 
of  time,  but  one  mile  an  hour  faster.     How  far  did  he  go? 

3.  If  the  distance  of  the  earth  from  the  sun  is  x  diameters 
of  the  earth,  the  distance  of  Jupiter  5  times  that  of  the  earth, 
the  distance  of  Saturn  twice  that  of  Jupiter,  and  the  distance 
of  Uranus  twice  that  of  Saturn,  what  is  the  sum  of  all  these 
distances? 

4.  A  charitable  association  working  six  consecutive  days 
collected  x  dollars  on  the  first  day,  and  on  each  of  the  remain- 
ing five  days  it  collected  a  dollars  more  than  on  the  day  pre- 
ceding. It  then  divided  the  amount  equally  among  ^x  white 
and  ha  colored  families.     How  much  did  each  family  get? 

5.  A  library  contains  ^x  books  of  history,  2y  books  less 
of  biography  than  of  history,  as  much  of  poetry  as  of  biog- 
raphy and  history  together,  and  Zy  fewer  novels  than  books 
of  poetry.  The  books  were  divided  equally  among  9  alcoves. 
How  many  were  in  each  alcove? 

6.  In  winding  up  a  company  each  stockholder  got  x  dollars, 
each  ordinary  creditor  got  {h  +  x)  dollars  more  than  each 
stockholder,  and  each  preferred  creditor  got  2a;  dollars  more 
than  each  ordinary  creditor.  There  were  in  all  a  stockholders, 
h  creditors  and  c  preferred  creditors.  What  was  the  total 
amount  divided? 

7.  A  railway  train  having  to  make  a  journey  of  600  miles 
ran  x  hours  at  the  rate  of  30  miles  an  hour  and  y  hours  at  the 
rate  of  45  miles  an  hour;  it  then  became  disabled  and  had  to 
make  the  remaining  distance  at  the  rate  of  15  miles  an  hour. 
How  long  did  it  require  to  make  the  entire  journey? 

Metliod  of  Solution.  We  must  subtract  from  the  totaJ  distance  the 
distances  it  ran  during  the  x  hours  and  the  y  hours.  The  remainder  is 
the  distance  it  had  to  run  at  the  rate  of  15  miles  an  hour.  Dividing 
this  remainder  by  15  will  give  the  time  required  for  the  last  stage 
of  the  journey.  Adding  the  times  of  the  first  two  stages,  which  are 
given,  we  shall  have  the  whole  time  required,  which  we  shall  find  to 
be  (40  —  ic  —  2y)  hours. 


42 


ALGEBRAIC  OPERATIONS. 


Addition.     < 


Memorakda   for  Review. 
Define  and  Explain:    Use  of  Parentheses;  Value  oi 
Symbol;  Power;  Exponent;  Degree;  Expression;  Ooejfficient; 
Term;  Monomial;  Binomial;  Trinomial;  Factor;  Equation. 

Addition  and  Subtraction. 
Define:  Like  Terms;  Algebraic  Addition  and   Subtrac- 
tion; Numerical  Addition  and  Subtraction;  Minuend;  Subtra- 
hend; Kemainder. 

When  the  terms  are  unlike. 
Rule  for  like  terms. 
Case  of  negative  terms, 
i    General  rule  for  positive  and  negative  terms. 

Subtraction.  Rule;  Reason  for  the  rule. 

Clearing  of    J   Sign  -|-  before  parentheses;  Rule. 
Parentheses.    (  Sign  —  before  parentheses;  Rule;  Reason. 

Multiplication. 
Define:  Multiplier;  Multiplicand;  Product. 

A  monomial  by  a  number;  Rule. 

One  monomial  by  another;  Rule. 

Like  symbols,  using  exponents;  Rule. 

More  than  two  factors;  Method;  Rule. 

A  polynomial  by  a  monomial;  Rule. 

A  negative   term  by  a   positive   multiplier; 

Principle. 
Any  term  by  a  negative  multiplier;  Rule. 

Divisiion. 
Dividend;  Divisor;  Quotient;  Exactly  Divisible; 
Prime;  Composite;  Common  Divisor;  Highest  Common  Di- 
visor; Multiple;  Common  Multiple;  Lowest  Common  Multiple. 
First  principle  of  division;  Rule  deduced. 
Rule  of  exponents. 
\  Division  of  polynomials;  Rule. 


Product  of 


Define ; 


Rules  and 
Principles. 


Rule  of  signs. 
Rule  for  factorinsf. 


Divisors  and  j   Highest  Common  Divisor;  Rule. 
IMultiples.      (  Lowest  Common  Multiple;  Rule. 


CHAPTER  III. 

ALGEBRAIC    FRACTIONS. 

56.  Def.  A  fraction  is  the  expression  of  an  indi- 
cated division  formed  by  writing  the  divisor  under  the 
dividend  with  a  line  between  them. 

Example.     The  quotient  oip  ^  q\%  the  fraction  — . 

The  numerator  of  the  fraction  is  the  dividend. 
The  denominator  is  the  divisor. 
Numerator  and  Denominator  are  called  terms  of 
the  fraction. 


SECTIOiq"   I.      MULTIPLICATIO]^   AIS^D   DlVISIOlS^   OF 

Fractions. 

57.    Theorem.     Multiplying  the  numerator  multiplies 
the  fraction. 

71% 

Reason.     If  we  call  a  fraction  — ,  it  will  be  the  same  as 

n 

m  wths.     If  we  multiply  it  by  any  factor,  as  3,  the  result  will 

be  3m  ;^ths,  that  is  ~^-. 
n 

EXERCISES. 

Multiply: 

1.   -  by  a.  3.    -  by  m. 

n    ^  n     ^ 

3.   —^  by  aV,  4.    —  by  nm. 

^  ^      \ 

5.   —  by^?^  6.   i-^  by  5<?. 

pq    ^  ^  X         '' 

-,    a(m-\-n)  ,     _.  r,     1  i       « 

7-  -^^  by  ^y-  8.   -^  by  a\ 

9.   — ^by9.  10.   ^-^hj4.xy. 


x  —  y    '  x-\-y 


44  ALGEBBAIU  FRACTIONS, 

n.   ^y  -  f  >'  by  9«J. 

ax  —  by 

12.   '  3   ,   ^  by  —  3w.       Ans. 3   ,   »    . 

13.  -::^by-i. 

14.   -^; — — ^  by  —  ahc. 

a-\-b  —  c      ^ 

Perform  the  indicated  multiplications: 

,„    ^   (a       am\  .  da"      da^m 

15.  Sai-r ).  Ans.     — 


c  J  bo 

\y       X 


16.    J^-^-l). 


2 

I 'in.  'r 

17.   ^a^b 


\n^       mj' 
18.   da'(--i-'^-^-\b'c\ 

fa  —  b  ,    Z>  —  c  ,   (?—  «\   „ 

.3.  -4^-^). 


V    c^       '       a'      '       b' 

.  am  —  an  ,   cm  —  en 

Ans.     • ; = . 


58.  Theokem.  Dividing  the  numerator  divides  the 
fraction. 

The  reason  is  nearly  the  same  as  in  the  case  of  multipli- 
cation. 

EXERCISES. 

Divide: 

,     2ax'  ,  ,  2x' 

1.   by  ax,         Ans.     — . 

mn     ''  mn 

2.  __5_by2^y.  3.   -^^^J^n^m. 


MULTIPLICATION  AND  DI  VISION  OF  FRACTIONS.      45 

.     aia  —  bY,  ,  ^    a(a  —  b)\         .,        . 

^'  77 — rrf  by  «  —  &.  5.   77 — — rf  by  —  (^  —  a). 

Note,     First  free  the  divisor  from  the  parentheses.     Coinp.  §33, 
which  shows  that  ~(b  —  d)  is  the  same  ^^—h-\- a  ox  a  —  h. 

6.  ^^^!£Lby7cW. 

^    3  .  4  .  5(«  +  J)  (c  +  ^)  ,     ^     .,     ,     ,, 

7.  17^/  by^.4(g  +  ^). 

8.  — ^-  by  -  «m%^  9.  ^(^qr^-  by  ^'  -  y'. 

10.  -T by  Sa, 

^^     aim  —  n)       c(m  —  n)  , 

11.  -^ — 5 ^^ — 7 — -  m  m  —  n. 

b  d  ^ 

13.  !^  +  l^by3.'. 

?  i' 

j3_   {a  +  h)(a-l)  _  (.-^)(«  +  i)  .     _  (J  _  „)^ 
m  n  ''       ^  ' 

14.  —  +  _-  +  _  by  «fc. 

15.  (l+<_(l  +  .)(l-«)  (  )^ 

mn  nr  j       \     *     / 

16.  ^ilzi.'!!)  +  Oi+IIO?Lzl)by«'-l. 

m  n  ^ 

^^    abjx^  +  y'  +  z')       Hf  +  ^^  +  ^•^)^  .    (^^  +  ^'  +  /)^^ 
c  a  b 

by  ^'  +  2/'  +  ^'. 
18.   -^__---^-^---by7«Jm^ 

19./^+~^^^~^ny/-^. 
«  b       ^  c  ^ 

^^-  r -  273 +273747 5  ^y-*^- 


46  ALGEBBAIC  FRACTIONS. 

59.  Theoeem.     Multiplying  the  denominator  divides  the 
fraction. 

Reason,    Suppose  a  line 
first  divided  into  4  parts.  |     '     '    1     '     '     |    '     '    1    '     '1 
One  of  the  parts  will  then 

be  -  of  the  line  and  n  of  them  will  then  be  -  of  it. 
4  4 

Now  multiply  the   denominator  4  by  3.      This  will   be 

dividing  the  line  into  3  .  4  =  12  parts,  so  that  each  fourth 

will  be  divided  into  3  parts.      Therefore  j  "^  ^  =  — ,  and 

n         _  n 
4""^-i2- 

EXERCISES. 
■1      ^'         o         A  ^'  ^      3« 

1.   _^3.     Ans.^.  2.  -^-^5. 

3.  ^^  ~-  5/^.  4.  i  -^  2n. 

5k  n 

5.  3-^-4(l-a).  6.  3^-2(1  +  ^). 

11.  -\  -  ^  -^  «Jc.  12.  J-  -  -i-  ^  3  .  4 .  5. 

ab      be  3.44.5 

13.  x'  -\-l-\-\~  x\     Ans.  1  +  -,  4-  -,. 

'   a;'  ^   x^      X 

14.  ^,+  --^3F.  15.  «_-+-_  J  ^a'. 

16. -2  -^  aa:^. 

2/        ^ 

17    _1 ^  _     ^      .   ^^ 

•*•'•      55         9         /-kQ  "»'  aXt 
ax       ax      2ax 


MULTIPLICATION  AND  DIVISION  OF  FRACTIONS.     47 

60.  Theorem.  Dividing  the  denominator  multiplies  the 
fraction. 

Reason.  If  the  fraction  is  ^,  and  we  divide  the  denomi- 
nator 13  by  3  we  have  j,  which  is  3  times  as  much  ^  j-^. 

^  ,        n 

So  also  £  is  3  times  as  much  as  ^^,  or 


EXERCISES. 


1.  |X2.  2.  |X3. 


6 
a 


3.|X6.  4.^jXl2. 

5.  (f^^).x  («  +  *)•  6.^x-^™'«- 

Note,    n  —  m  =  —  (m  —  n). 

^•I(3^W)^'-  ^'•^^(*-"^- 

20.  -^^i;X(:^  +  2/)-  21.  ;;.^Xa  +  J. 

23      ^  +  g    X  («'  + 1).  23.  ^-^  X  *. 
■*     a'm -\-m     ^    ^   '  ax-bx 


48  ALGEBRAIC  FRACTIONS. 

61.  If  we  multiply  a  fraction  —  by  its  denominator  q  we 

haye  by  the  preceding  rule^,   that   is  jt?,    as   the   product. 

Therefore 

A  fraction  is  multiplied  by  its  denominator  ty  simply  re- 
moving it. 

Examples.  -x2  =  1;  -x«^  =  2; ^x(l— F)=m. 

it  a  \.  —  Ic 

63.  When  the  multiplier  and  the  denominator 
contain  common  factors,  we  may  multiply  by  these 
factors  by  removing  them  from  the  denominator,  and 
then  multiply  the  remaining  factors  into  the  numer- 
ator. 

Example.     Multiply  —  by  mx. 

^  ''  mn    "^ 

We  multiply  by  m  by  removing  it  from  the  denominator, 
and  then  multiply  the  numerator  by  x.     Therefore 

a  ax 

— .  X  mx  =  — . 
mn  n 


EXERCISES. 

6.  j-3-  X  («  -  1)  {a  +  1).     7.  ^,-  X  U^'2/'. 
10-   (,_^)  (,!»)(,_,)  X  s{s  -  a). 


REDUCTION.  49 


"•    (l_^)(l_^-)X-a(g-l)(l-^). 


Section  II.    Reductiois-  of  Fractions. 

63.  Def.  Reduction  means  changing  the  form  of 
an  expression  without  changing  the  value. 

64.  Reduction  to  lowest  terms. 

Theorem.  //  loth  terms  of  a  fraction  le  multiplied  or 
divided  hy  the  same  quantity,  tlie  value  of  the  fraction  will 
not  he  altered. 

Reason.  By  §§  57,  58,  59  and  60  the  multiplications  of 
the  two  terms  have  opposite  effects,  which  cancel  each  other, 
and  so  have  the  divisions.  Hence  all  coiannon  factors  m  the 
two  terms  may  be  cancelled. 

Note.     Observe  that  only  factors  can  be  cancelled. 

G5.  Def.  When  all  the  common  factors  are  cancelled  the 
fraction  is  said  to  be  reduced  to  its  lowest  terms. 

To  reduce  a  fraction  to  its  lowest  terms: 

EuLE.     Factor  each  term  when  necessary,  and  cancel  all 

factors  common  to  both, 

^  ,  mnp^x^       mn 

Example  1.  — ^7-  =  — . 

rsp  X         rs 

The  factors  p^x^,  common  to  both  terms,  are  cancelled. 

x^y  _x 

xY    y 

the  factors  x'^y  being  cancelled. 

mx  —  nx  _(m  —  n)x  __  x 


Ex.  3. 


my  —  ny      {m  —  n)y      y 


EXERCISES. 


Reduce  to  lowest  terms: 

a¥c'  2    ah  +  ac* 

•^-  a'Vc  c'a  -  aV 


50  ALGEBRAIC  FRACTIONS. 

'     n'x{i  -  a)  '  '  i?Y+i^V+i?V* 

(l-xYJl  +  ^y  .    (a-i){a-2) 

(1  -x)  {x -{-!)'  (1  -a)  (2-  a)' 
Note,    a  — 1  =  —  {1  —  a). 

ISa^b^c*  amx  —  amy 

TZa^h^cxy^ '  '  m^c^x  -|-  ma^y 

rs  -  rV  /V+/y 

11.   (!-«'  +  a')^y  i3_      7.9  +  3   _ 

a:?/  +  a*xy     '  '7.9  —  2.3* 

3a^{k  —  q)xy  '  abc      abc      abc 

a;?/- "^  y;2- "^  ;2a;-*  3.  4.  5  "^  3.  4.  5  "^  3  .  4.  5* 

66.  Reduction  to  Given  Denominator.  An  entire 
quantity  may  be  expressed  as  a  fraction  with  any  re- 
quired denominator,  Z>,  by  supposing  it  to  have  the 
denominator  1  and  then  multiplying  both  terms  by  D. 

For  if  a  is  any  entire  quantity,  we  have 
a      aD 

''  =  i='B- 

Example.  If  we  wish  to  express  the  quantity  ai  as  a 
fraction  having  xy  for  its  denominator^  we  write 

adxy 
xy  * 

67.  If  the  quantity  is  fractional,  both  terms  of  the 
fraction  must  be  multiplied  by  that  factor  which  will 
produce  the  required  denominator. 

Example.      To  express  j-  with  the  denominator  nd^  we 

multiply  both  members  by  nb^  -^  h  =  nb^.     Thus, 

a  _  anb^ 

b~''W 

This  process  is  the  reverse  of  reducing  to  lowest  terms. 


REDUCTION.  51 


EXERCISES. 

Express  the  quantity: 
1.  m  with  denominator  a.  Ans.  — . 


2.  m  with  denominator  m.  Ans.   — . 

Z.  a  —  b  with  denominator  x. 


a 
m 


4.  -  with  denominator  or, 
Q 

5.  —  with  denominator  12. 
4 

6.  a^  with  denominator  a  -\-  b. 

7.  X  —  y  with  denominator  x  -\-  y. 

1  —  k^ 

8.  with  denominator  abc. 

a 

9.  ^ — Jl —  with  denominator  lOaa;^ (F  —  1), 

10.  — -^^ with  denominator  20m^ 

—  4m 

1  —  ^' 

11.  ^  with  denominator  1  —  a. 

9  -^ 
Note.    1  —  ^^  =  —  (^  —  1). 

12.  :r-,     —     and     — :  with  denominator  abc» 
be       ca  ab 

13.  — ,     —     and     —  with  denominator  rst, 
r       s  t 

X^  |/2 

14.  —J,     j-^     and    1  with  denominator  a^b^. 

s  s  s 

15.   ,    i     and    with  den.  {s—a)(s—b)(s—c). 

s  —  a     s  —  b  s  —  c  ^        ^^        ^^        ' 

16.  with  denominator  +  a. 

—  a 

ct  X 

17.  T-     and     -  with  denominator  abxy, 
by  ^ 


52  ALGEBBAIG  FRACTIONS. 

68.  To  reduce  fractions  to  a  common  denominator: 

Rule.      Choose  a  common  muUijjle  of  the  denominators. 

Multiply  both  terms  of  each  fraction  by  the  multiplier 
7iecessary  to  change  its  denominator  to  the  chosen  multiple. 

Note  1.  Any  common  multiple  of  the  denominators  may 
be  taken  as  the  common  denominator,  but  the  L.  0.  M.  is 
the  simplest. 

Note  2.  The  required  multipliers  v/ill  be  the  quotients 
found  by  dividing  the  chosen  multiple  by  the  denominator  of 
each  separate  fraction. 

Note  3.  When  the  denominators  have  no  common  factors, 
the  multiplier  for  each  fraction  will  be  the  product  of  the 
denominators  of  all  the  other  fractions. 

Note  4.  An  entire  quantity  must  be  regarded  as  having 
the  denominator  1.     (§  66.) 

Example  1.     Eeduce  to  common  denominator 

\  T)  ,         m 

w,        — ,        -^  and        — . 

m         mn  np 

The  given  denominators  are  1,  m,  mn,  np. 

Their  L.  0.  M.  is  mnp. 

The  multipliers  are  (Note  2)  mnp,  np,  p  and  m. 

Multiplying  by  these  quantities,  the  fractions  become 

m^np  np  p^  ^  m^ 

-,        —^—,        -^-—         and        . 

mnp  mnp         mnp 

Ex.3.  \        I        i. 

a  b  c 

By  Note  3  the  multipliers  are  be,  ac  and  ab. 

Multiplying  by  them,  the  fractions  become 

bcx         acy  _        abz 

—Ti        —r         and        -^. 
abc         abc  dbo 

EXERCISES. 

Reduce  to  common  denominator: 

1    1      -      -  2 

b       c      a 
abc 


5. 


X         3 


1 

1       1 

a' 

y   T 

m" 
a'' 

m 

~,    m. 
a 

2x 
a' 

3^         42 
1  -  «'     a' 

AGGREGATION.  53 

q    ^      ^      ?  10    i         ^        1 

^•3'     V     5*  3?      4/      bz 

M     ^      ^      £1  19    A  m         5n_ 

■f    -r         ,.  .   -a' 


13.  /;  — ,    ^^^^  .  14.  «^,  ^-^,   ^345. 

15.  -, ,     — .  Id. 


c 


m—V     1  —  7n'     m  '   1  —  x'     2a;(l  —  x)'      Sx"^' 


Section  III.    Aggregation^  and  Dissection  of 
Fractions. 

69.  I>ef,  Aggregation  is  the  expression  of  the 
algebraic  sum  of  several  fractions  as  a  single  fraction. 

Def.  Dissection  is  the  separation  of  a  fraction 
into  an  algebraic  sum  of  fractions. 

70.  Case  I.  When  the  fractions  to  be  aggregated  have 
the  same  denominator. 

EuLE.  1.  Unclose  each  numerator  detween  parentheses,  or 
suppose  it  so  enclosed. 

2.  Prefix  to  each  numerator  so  enclosed  the  algebraic  sign 
of  the  fraction. 

3.  Form  the  algebraic  sum  of  the  expressions  thus  found 
and  write  the  cofnmon  denominator  under  them. 

Reason.  1.  By  the  definition  of  a  fraction,  the  numerator 
expresses  a  number  of  fractional  units. 

2.  Hence  the  sum  of  several  fractions  with  a  common  de- 
nominator is  the  sum  of  all  the  fractional  units  indicated  by 
the  several  numerators. 

3.  If  the  fractions  are  preceded  by  the  minus  sign,  it  in- 
dicates that  the  fractional  units  in  its  numerator  are  to  be 
algebraically  subtracted.  This  subtraction  is  indicated  by 
enclosing  the  numerator  between  parentheses  and  prefixing 
the  minus  sign. 

4.  Hence  the  algebraic  sum  of  all  the  fractions  is  formed 
in  the  manner  directed  in  the  rule. 


54  ALGEBRAIC  FRACTIONS. 

Example  1.     ;/  ~  ^  +  j  means  l  fractional  units  to  be 

subtracted  from  a  fractional  units,  and  c  fractional  units  to 
be  added.     Hence 

a___i       c  _a  —  'b-\-c 

Ex.  2. 

c^^x     -3g  —  %x  __  4a; -8c  _  c-^hx  —  {^c  —  2a;)  — (4a;  -  8c) 

D  ~  D  D D 

_  c  4-  5a;  —  3c  +  2a;  —  4a;  +  8c  . 
_  _ 

_  6c  +  3a; 
~"       D      • 


Aggregate: 


EXERCISES. 


^    a  —  X  ^   a-\-x  .          %a 

1. ' — .  Ans.     — . 

mm  m 

_«  +  a;      a  —  X  .           a-\-x—{a  —  x\      2x 

2.  — ' .  Ans.     — ■ -5^ ^  =  — . 

mm  mm 

«  +  2a;       a  ~  X  2a  —  3x      a  -\-'Sx 

m              m    '  '        m               m 

^     a  -\-  h       2a  —  dd  ^        y              x 

5.  ■ .  6.  -  ^ 


m  —  n       m  —  n  ^  —  y      ^  —  V 

Sx  3a  a  —  2b-]-Sc      a-{-2b~Sc 

'  X  -{-  a      X  -\-  a'  '       1  —  x^  1  —  x^     ' 

s  s  s     '         '       m  m    ' 

When  a+  5  +  c  =  2s,  what  does  the  answer  to  Ex.  9  reduce  to? 

4:xy^         4:xy^         Axy^  ' 

,_    a^-{-2aI)-\-i'      a""  -  2ah -\- V 
12. 


mn  mn 

^_    m{\  -  a)   ■  m{a  -  1)  2(1  -  F)       1-F 

If      ^        ¥       '  2.3.4        2.3.4' 

2a -^l      2a  ^b      5a -2b  a-}- 2b 

■^^'      D             D     '^      D  '  D    ' 


AGOBEQATION.  55 

71.  Case  II.  If  an  entire  quantity  and  a  fraction 
are  to  be  aggregated,  we  reduce  the  entire  quantity 
to  a  fraction  having  the  same  denominator  as  the 
fraction. 

Example.     Aggregate  a-\ — . 

Til 

CLYh 

Eeducing  by  §  66,  we  have  a  =  — .     So  the  sum  is 

an      m  _an  -\-  m 
n        n~       n 
We  now  see  that  the  result  is  obtained  by  the  following 
rule: 

Multiply  the  entire  quantity  hy  the  denominator,  and  add 
the  product  with  the  numerators. 

EXERCISES. 

Aggregate: 

^     ^  ^  A  1  —  x  —  a 

1.  1  — .         Ans.     — . 

1  —  x  1  —  x. 

2.  l+T-^-.  3.   l-:rT-- 

'  1  —  a;  l-\-x 

6./+}.  7.  .  +  3,-1-^:^^. 

8.   1  —  r-r — .  9.  Amn  —  -^    a  ,. 

1  -\-  a  4rm^n* 

10.  x-j-^^.  11.  A  ^ 


a:  +  2*  4  +  5a'' 

12.   m-h  +  -,  13.   4:Aa  ^ 


m'  '  a-\-  A' 

14.   .  +  1^.  15.   .  +  i. 

16.   x^a-^-.  17.   l^x--^-=^ 


18. 


a-* 

a'  4-  2a^  +  ^'  _    4:ah        a'-2ab-^b* 
Samx'  3amx'  Zamx' 


5e  ALGEBUAIC  FRACTIONS. 

*72.  Case  III.  When  the  fractions  have  different 
denominators  we  reduce  them  to  a  common  denomina- 
tor and  proceed  as  in  §  70. 


EXERCISES. 

Aggregate: 

■•  ,-+r 

Ans. 

a-\-b 
ab  ' 

2. 

1       1 

b       a 

3   !?.l_!?L 

4. 

m          n 

'a        I' 

m  —  n      n 

5        "^      -4- 

n 

6. 

a          3b 

O,                        — r- 

m-\-n 

m 

a  —  b       a' 

^    a       h 

c 

x'       x^ 

'•  ^-+3  + 

4* 

8. 

^  +  7  +  ^- 

1               1 

1 

a  —  b      b  —  c      c  —  a 

ab      be 

+-.«• 

10. 

c      '      a      '       b    ' 

n.i  +  i  + 

c    '   a 

12. 

^^\^W,- 

13.  l-f  + 

r 

14. 

.   /'  ,    /' 

2.3. 

4' 

•'       2.3  '  a.  3.  4.  5" 

1 

1        1 

15.    1-4- 

16. 

^'-W  +  '- 

.^    1       1 

1                1 

i^-     — 2e' 

18. 

rrf      wr 

m'      m^''' 

73.  Dissection  of  Fractions.  If  the  numerator 
is  a  polynomial,  each  of  its  terms  may  be  divided 
separately  by  the  denominator,  and  the  several  frac- 
tions connected  by  the  signs  +  or  — . 

The  principle  is  that  on  which  the  division  of  polynomials 
^  founded  (§  46).     The  general  form  is 

±i:^iL£±^  =  £  +  ^  +  £  +  etc.        (1) 

m  m      m      m 

The  separate  fractions  may  then  be  reduced  to  their  lowest 
terms. 

Example.     Dissect  the  fraction 

12^^g  +  Qab  —  3g 
UabG  ' 


MULTIPLICA  TION  AND  DIVISION  B  T  FBA  CTI0N8.       57 
The  general  form  (1)  gives  for  the  separate  fractions 


12abc       VZabc       12abc 
Reducing  each  fraction  to  its  lowest  terms  the  sum  becomes 

1  +  1— i-. 

EXERCISES. 

a*  —  a;'  +  2ax  x^  —  2xy  -f-  y* 

ab-\-bc-{-ca  .    am^  —  2amn  —  «' J' 

O.  7  »  4:.  2      a  • 

abc  aw; 

Note,    f—g  and  h—  j  are  to  be  considered  as  single  quantities;  for 
example,  like  A  and  B  in  Ex.  5. 

aY  +  ^'^'  -  ^'^'  o    1  +  <?  +  <^'  +  g' 

y  +  g'  +  r'  3'  +  4'  +  5' 

2jt?^r       •  2.3.4.5  * 

x^  -2px-q  a{l  -  x)  -  b(l  +  x) 

J.1.     J- .  14). 7 . 

2pqx  ab 

a{l  -  x)  -  b{l  +  x)  a'^-b'' 

{l-x){l-\-x)    •  ^^'      aH-  ' 

a^^W""  '  A"B*'*' 


Section  IY.    Multiplication  and  Division  by 
Fractions. 

74.  Multiplication  by  Fractions. 

Def.  To  multiply  a  quantity  by  a  fraction  means 
to  take  the  same  part  of  the  quantity  that  the  frac- 
tion is  of  unity. 

Example.     To  multiply  Q  by  —  means  to  take  —  of  Q, 


58  ALGEBRAIC  FRACTIONS. 

To  form  —  of  Q,  we  divide  Q  into  n  parts,  and  take  m  of 

those  parts.     Hence: 

To  multiply  hy  a  fraction,  we  divide  the  multiplicand  ty 
the  denominator  and  multiply  the  result  ly  the  numerator, 

EXERCISES. 

Multiply: 

1.  a  +  ^by^.         Ans.     -+_  =  --+a. 

2.  «-^by^.  3.  i.»  +  ^'by|. 

1  77i 

4.  m  +  w  —  1  by  — .  5.  m  —  7i  +  1  by  — . 

^  mn  '        -^  7i 

6.  a^h  -  aV  by  -  4-  7.  a  +  a'  +  «^  by  i . 

8.  4a:y  +  7«ir' -  11J«/ by  — . 

9.  a^_JFby-^. 

10.  «"  +  «»»  +  fl^'»  by  ^, 
a 

a 


11.  a{f-g)-l\f-g)  ^7  ifjzrfy 
13.  (1  -  a.)b  -  a{l  -  h)  by 


(1 -«)(!-*) 

14.  1  —  a;  —  V  —  ^  by . 

^  "^  xyz 

15.  2a»  +  3J»by||i-. 

75.  The  Multiplicand  a  Fraction.     If  the  multiplicand 

is  also  a  fraction,  suppose  j-,  and  is  to  be  multiplied  by  — , 

then  by  §§  57, 59,  we  multiply  by  w  by  multiplying  the  numer- 
ator, a,  and  divide  by  n  by  multiplying  the  denominator,  b. 


MULTIPLICATION  AND  DIVISION  BY  FRACTIONS.      69 

„  am      am 

Hence  t-  X  —  =  -r — . 

0       n       on 

That  is,  we  multiply  the  numerators  for  a  new  numerator, 
a7id  the  denominators  for  a  new  denominator. 

The  result  should  then  be  reduced  to  lowest  terms. 

EXERCISES. 

Execute  the  following  multiplications: 

^     a       mb  .  amb       m 

1.  T-  X  — .         Ans.     — ~  =  — . 
0       pa  apo       p 

gx^      3^'a;  ^    2am^x  ^  2my 


pq         gr  *  '     Zhy         Shx' 

U  ^  aji 


.    a  —  b      m—  n  ir/^,^\^ 

m  a  —0  \b       a  JO 


P      P"W  ^    /'^    ,   ^'    ,  ^'Y 


6       {^_^)l..  7.     f^-L^-L^^ 

\^       q^Jp"^  \r       r""       r^Jc 

8. — - —  — - — .  9.   [a' -\- ac -\- -r-]-. 

\l  —  r       l-\-rjl-\-r  \                   b  Ja 

,^    m      ^V   ,   IbbcVZa  -,,     /     .      ^  ,  ^  y'\x 


10/  a^       a^X  1 


3  ^  5 


/I  +  «  y^*    \mw(l  —  a) 

'   \  m^         1  —  aj     1  -{•  a 


14    f^!_^^^  15    /^  ,  «y    ,   «i.\^ 


xy  I  a 

76.  Division  hy  Fractions. 

Def.  To  divide  a  quantity  by  a  fraction  means  to 
find  that  expression  which,  when  multiplied  by  the 
fraction,  will  produce  the  quantity. 

If  the  dividend  is  7-  and  the  divisor  is  — ,  then  the  quo- 
b  n  ^ 

...  .  .     na    .  na       m      a      ^^  .      , .  .  - 

tient  must  be  — 7  because  — -  x  —  =  t--     Hence,  to  divide 
mb  mb      n       b 

by  a  fraction, 

Invert  the  terms  of  the  divisor  and  multiply  by  the  frac- 
tion thus  formed. 


eO  ALOEBRAIG  FJRAGTIONa. 


EXERCISES. 

Execute  the  following  diyisions: 
^     r    ,  r                              c.    ^r  ^    r 
s       s                                     s       2s 

3.  a  -^  — .                            4.  m  -7-  — . 
a                                          m 

5.  1  :  f . 
1      P 

o  p'  .  q" 
8-   y.   •  ^. 

4aw£^_^4«w  /^^  _L. -"^ -i- 2. 

5Z>?i?/^    *    5^/i *  '    \b  ~^  c I   '  c' 

11.  (4_^«]^?!|!.  13.  (_!!!_ _l+£)^l±« 

W      y  J       «  Vl  —  a         n    J      1  —  a 

13.   (^  +  ^)^ 


{s  —  a)  (s  —  b)' 

14.  (i  _  1)  ^  -J_.      15.  (ir + ri + il)  -.  Jl. 

\        a  J      1—a  \s        a       sqj      grs 


l'7a'b'xy"'    ^  61m''n'x  13      ,      14 

*      19m/i2;     *    19a%^*  a— 1  *  1  —  a 

^^-  9    •        9a  ^^'  af^'^a^' 

20.    fl--lU4.  21.  1-^ 


Note.     The  answers  to  Exercises  5  and  21  show  that  the  quotient  of 
unity  divided  by  any  fraction  is  equal  to  the  fraction  inverted. 

MISCELLANEOUS    EXERCISES. 

Express  and  reduce: 

1.  I  of  I  of  Hx.  2-  T  ^f  I  of  ^^- 

3        5  o       z 

3.  I  of  I  of  {5x  -  a).  4.  J-  of  (|x  -  «). 

5.  lof(^*+nA  6.  lof'^of^ofx. 

m       \n     ^      j  m       n       p 

7.  From  a  sum  of  a:  dollars  -  of  the  amount  and  a  dollar 

o 

more  were  taken.     Express  —  the  remainder. 

o 


MEMORANDA  FOR  REVIEW.  61 

8.  Increase  the  quantity  xhj  —  part  of  itself,  and  express 

the  result  as  a  fraction,  with  ic  as  a  factor. 

9.  Diminish  the  quantity  x  by  the  ni\i  part  of  itself,  and 
express  the  result  in  the  same  way. 

10.  From  a  cistern  containing  g  gallons  of  water  -  the 

o 

water  was  taken  one  day  and  ^  of  what  was  left  the  next  day. 

Express  the  remainder. 

11.  A  father  left  h  thousand  dollars  to  each  of  a  children. 
Each  of  these  children  had  h  other  children  between  whom 
the  money  was  equally  divided.  How  much  did  each  grand- 
child get? 

12.  A  charitable  association  collected  x  dollars  from  each 
of  a  people  and  y  dimes  from  each  of  h  people.  It  divided 
the  amount  equally  among  a  almoners,  and  each  of  these 
almoners  divided  his  share  among  h  poor.  How  much  did 
each  poor  person  get? 

13.  A  man  having  to  make  a  journey  of  m  miles  went  — 

the  distance  and  a  miles  more  on  the  first  day,  and  —  the  re- 
maining distance  -f  ^  miles  on  the  second  day.  How  far  had 
he  still  to  go? 

14.  A  huckster  with  t  turkeys  sold  half  of  them  at  —  dollars 

1  m" 

each,  and  -  the  remainder  at  —^  dollars  each,  and  what  were 
3  n 

m' 
left  at  —5  dollars  each.     How  much  did  he  realize? 
n 


Memokakda  for  Review. 

The  Two  Conceptions  of  a  Fraction. 

1.  As  parts  of  a  unit. 

2.  As  the  quotient  of  an  indicated  division. 

Explain  significance  of  numerator  and  denominator  in 
each  mode  of  conception. 


ALGEBRAIC  FBACTIONS. 


Multiplica 
tion 


Division 


Multiplication  and  Division  of  Fractions, 
"  By  multiplying  numerator,-  Reason. 
By  dividing  denominator;  Eeason. 
By  removing  denominator;  Reason. 
When   multiplier  and  denominator  have  a 
common  factor. 

By  dividing  numerator;  Explain. 
By  multiplying  denominator;  Explain. 
When  divisor  and  numerator  have  a  common 
factor. 


Define 


Reduction 


Aggrega- 
tion 


Dissection. 


Reduction  of  Fractions, 

Reduction;  Lowest  Terms. 

'  By  multiplying  both  terms  by  the  same  fac- 
tor; Explain. 
By  dividing   both  terms   by  the  same   di- 
visor; Explain. 
To  lowest  terms. 

(  Entire  quan- 
To  given  denominator;  Reason.  •<     tity. 

(  Fraction. 
To  common  denominator. 


Aggregation  and  Dissection, 

Of  fractions  having  the  same  denominator; 

Give  rule  and  reasons. 
Of  an  entire  quantity  and  a  fraction. 
Of  fractions  having  different  denominators. 
Show  when  applicable;  Give  rule;  reason. 


Multiplication  and  Division  iy  Fractions, 


Multiplica- 
tion. 


Division. 


Define  multiplication  by  a  fraction. 
Give  general  rule;  Explain  reason. 
Show  how  general  rule  is  applied  when  the 
multiplicand  is  also  a  fraction. 

Define  division  by  a  fraction. 
Deduce  general  rule. 


CHAPTER  IV. 

SIMPLE     EQUATIONS. 


Definitions. 

77.  Bef.  An  equation  is  a  statement,  in  the  lan- 
guage of  algebra,  that  two  expressions  are  equal. 

Def.  The  two  equal  expressions  are  called  members 
of  the  equation. 

78.  Def.  An  identical  equation  is  one  which  is 
true  for  all  values  of  the  algebraic  symbols  which  enter 
into  it,  or  which  has  numbers  only  for  its  members. 

Examples.     The  equations 

14  +  9  =  29  -  6, 
(5  +  13)  -  (3  X  4)  -  6  =  0, 
which  contain  no  algebraic  symbols,  are  identical  equations. 
So  also  are  the  equations 

X     =     Xy 

X  —  X  =  0, 
^x  -f  y)  =  72;  +  ly, 
because  they  are  necessarily  true,  whatever  values  we  assign 
to  X  and  y. 

Remark.  All  the  equations  used  in  the  preceding  chapters  to  express 
the  relations  of  algebraic  quantities  are  identical  ones,  because  they  are 
true  for  all  values  of  these  quantities. 

79.  Def.  An  equation  of  condition  is  one  which 
can  be  true  only  when  the  algebraic  symbols  are  equal 
to  certain  quantities,  or  have  certain  relations  among 
themselves. 

Example.     The  equation 

a:  +  6  =  22 
can  be  true  only  when  x  is  equal  to  16,  and  is  therefore  an 
equation  of  condition. 

Remark.  In  an  equation  of  condition,  pome  of  the  quantities  may 
be  supposed  to  be  known  and  others  to  he  unknown. 


64  SIMPLE  EQUATIONS. 

80.  Def.  To  solve  an  equation  means  to  find  such 
numbers  or  algebraic  expressions  as,  being  substi- 
tuted for  the  unknown  quantity,  will  render  the  equa- 
tion  identically  true. 

Any  such  value  of  the  unknown  quantity  is  called 
a  root  of  the  equation. 

Example  1.     The  number  3  is  a  root  of  the  equation 
W  -  18  =  0, 
because  when  we  put  3  in  place  of  x  the  equation  is  satisfied 
identically.     Prove  this.     The  number   —  3  is  also  a  root. 

An  algebraic  equation  is  solved  by  performing 
such  similar  operations  upon  its  two  members  that  the 
unknown  quantity  shall  finally  stand  alone  as  one 
member  of  an  equation. 

Remark.  It  is  common  in  Elementary  Algebra  to  represent 
unknown  quantities  by  the  last  letters  of  tlie  alphabet,  and  quantities 
supposed  to  be  known  by  the  first  letters.  But  this  is  not  at  all  neces- 
sary, and  the  student  should  accustom  himself  to  regard  any  symbol  as 
an  unknown  quantity. 

Axioms. 

81.  Def.  An  axiom  is  a  proposition  which  is 
taken  for  granted,  in  order  that  we  may,  by  it,  prove 
some  other  proposition. 

Equations  are  solved  by  operations  founded  upon  the  fol- 
lowing axioms,  which  are  self-evident,  and  so  need  no  proof. 

Ax.  I.  If  equal  quantities  be  added  to  the  two 
members  of  an  equation,  the  members  will  still  be 
equal. 

Ax.  II.  If  equal  quantities  be  subtracted  from  the 
two  members  of  an  equation,  they  will  still  be  equal. 

Ax.  III.  If  the  two  members  be  multiplied  by  equal 
factors,  they  will  still  be  equal. 

Ax.  IV.  If  the  two  members  be  divided  by  equal 
divisors  (the  divisors  being  different  from  zero),  they 
will  still  be  equal. 

Ax.  Y.    Similar  roots  of  the  two  members  are  equal. 

These  axioms  may  be  summed  up  in  the  single  one, 

Similar  operations  upo7i  equal  quantities  give  equal  results. 


TRANSPOSING  TERMS.  66 

Transposing  Terms. 

82.  Theorem.  Any  term  may  he  transposed  from  one 
member  of  an  equation  to  the  other  member  if  its  sign  be 
changed. 

Example  1.    From  the  equation 
7  +  18  =  25 
we  obtain,  by  transposing  18, 

7  =  25  -  18; 
by  transposing  7, 

18  =  25  -  7. 
Ex.  2.     From  the  equation 

9  =  12  -  3 
we  obtain,  by  transposing  3, 

9  +  3  =  12, 
and  from  this  last  equation,  by  transposing  9, 
3  =  12  -  9. 

EXERCISES. 

Form  two  equations  from  each  of  the  following  by  trans- 
position: 

1.  16  =  9  +  7.  2.  8  +  5=13. 

3.  15  =  6  +  9.  4.  23  -  10  =  13. 

5.  14  =  20  -  6.         Ans.  14  +  6  =  20;  6  =  20  -  14. 

6.  14  =  21  -  7.  7.  17  -  8  =  9. 
Form  as  many  equations  as  you  can  from: 

8.  8  -  5  =  9  -  6.  9.  19  -  7  =  15  -  3. 

83.  General  Proof  of  Transposition.  Let  us  have  the 
equation 

a-\-t  =  b. 

Now  subtract  t  from  both  members, 

a-\-t  —  t  =  b  —  t; 
whence,  by  reduction,  a  =  b  —  t. 

This  equation  is  the  same  as  the  one  from  which  we  started, 
except  that  t  has  been  transposed  to  the  second  member,  with 
its  sign  changed  from  +  to  — 


QQ  SIMPLE  EQUATIONS. 

If  the  equation  is 

i  —  t  =K  a, 

we  may  add  t  to  both  members,  which  would  giye 

EXERCISES  IN   SOLVINQ  EQUATIONS   BY  TRANSPOSITION, 

1.  Find  that  number  which,  when  9  is  added  to  it,  will 
make  25. 

Solution.    Let  us  call  the  required  number  n.     The  prob- 
lem says  9  must  be  added  to  it.    Adding  9  the  sum  is  w  +  9. 
The  problem  says  this  sum  must  make  25.     Therefore 

^-f-9  =  25. 
Transposing  9, 

w  =  25  — 9  =  16.  Ans. 

Note.  This  and  some  of  the  following  problems  are  so  simple  that 
the  pupil  can  answer  them  mentally;  but  he  should  do  them  by  algebra 
in  order  to  learn  methods  which  may  be  applied  to  more  diflBcult  prob- 
lems. 

2.  Find  that  number  which,  when  17  is  added  to  it,  will 
make  30. 

3.  Find  that  number  which,  when  12  is  subtracted  from 
it,  will  leave  the  remainder  11. 

4.  Find  a  number  which,  subtracted  from  16,  will  give  9 
as  the  remainder. 

Solution.  Let  x  be  the  number.  When  subtracted  from 
16,  the  remainder  is  16  —  a:.  By  the  conditions  of  the  ques- 
tion, 

16  -  a;  =  9. 
Transposing  x,  we  have 

16  —  9  -f  a;. 
Transposing  9, 

U-9=x, 
whence  x  =  7.    Ans. 

5.  Find  a  number  which  being  subtracted  from  15,  the 
remainder  shall  be  6. 

6.  If  a  number  be  diminished  by  6,  and  the  remainder 
multiplied  by  3,  the  product  shall  be  double  the  number. 
What  is  the  number? 

Solution.  Let  r  be  the  number.  Diminishing  it  by  6 
the  remainder  is  r  —  Q.     Multiplying  this  remainder  by  3 


EQUATIONS  SOLVED  BT  TRANSPOSITION,  67 

the  product  is  3r  —  18  (§38).      By  the  condition  of  the 
problem  this  product  is  equal  to  2r.     Therefore 

3r  -  18  =  2r. 
Transposing  18, 

3r  =  2r  +  18. 
Transposing  2r, 

3r  —  2r  =  18,  , 

or,  by  reduction, 

r  =  18.  Ans.  ^ 

Proof,     18  -  6  =  12;  12  X  3  =  36,  which  is  twice  18. 

Note.  The  student  should  prove  all  his  answers  by  showing  that 
they  fulfil  the  conditions  of  the  problem. 

7.  If  4  be  subtracted  from  a  number,  and  the  remainder 
multiplied  by  4,  the  product  will  be  three  times  the  number. 
Find  the  number. 

8.  A  baker  started  out  with  x  loaves  of  bread.  He  sold 
all  but  8  of  them  at  '5  cents  each,  and  then  had  as  much 
money  as  if  he  had  sold  them  all  at  4  cents  each.  What  was 
his  number  ic? 

9.  One  baker  started  out  with  y  loaves,  and  another  with 
9  loaves  less.  The  first  sold  all  his  at  5  cents  each,  and  the 
other  sold  all  his  at  6  cents  each  and  realized  an  equal  amount. 
How  many  loaves  had  each? 

10.  A  huckster  bought  a  lot  of  turkeys  at  $1  each.  Nine 
were  spoilt,  and  he  sold  the  remainder  at  $2  each  and  made  a 
profit  of  $7.     How  many  did  he  buy? 

Solution.  Let  x  be  the  number  he  bought.  At  $1  each 
they  cost  him  x  dollars.  9  being  spoilt  he  had  ic  —  9  to  sell. 
At  $2  each  the  amount  was  2a:  —  18  dollars.  Because  he 
made  a  profit  of  $7  this  amount  must,  by  the  conditions,  be 
17  more  than  the  cost;  that  is  $7  more  than  %x.     Therefore 

2a;  -  18  =  a;  +  7. 
Transposing  x  and  18, 

22;  -  a;  =  7  +  18, 
or  x  =  25.  Ans. 

Proof.  25  —  9  =  16;  16  X  2  =  32,  which  is  the  same  as 
25  +  7. 


68  SIMPLE  EQUATIONS. 

Division  of  Equations. 

84.  From  the  equation 

12  -  8  =  4 
we  obtain,  by  dividing  by  2, 

6-4  =  2, 
and,  by  dividing  by  2, 

3-2  =  1. 

EXERCISES. 

Form  as  many  more  equations  as  you  can  from  the  follow- 
ing by  division,  and  see  if  the  results  are  true: 

1.  24  -  16  =  8.  2.  36  -  24  =  12. 

3.  72-45  =  27.  4.  84-70  =  35-21. 

85.  From  the  equation 

^x  =  24 
we  obtain,  by  dividing  by  3, 

a;  =  8. 
Proof,    Putting  8  for  x  m  the  given  equation,  we  have 
3  .  8  =  24, 
which  equation  is  identically  true. 
If  we  have  the  equation 

ax  =  m, 
we  find,  by  dividing  by  a, 

m 

x=  —, 

a 

Hence,  to  solve  an  equation  in  which  one  member  is 
known  and  the  other  member  is  the  unknown  quantity  mul- 
tiplied by  a  coefficient: 

EuLE.  Divide  loth  members  by  the  coefficient  of  the  un- 
known quantity, 

EXERCISES. 

Solve  the  following  equations  by  reduction  and  transposi- 
tion: 

1.  a-\-x  =  b  —2x, 

Solution,     Transposing  a  and  2x, 
x-\-2x  —  b  —  a, 
ox  3x  =  b  —  a, 


DIVISION  OF  EQUATIONS. 


Dividing  by  3, 

x  = 

5- 
3 

■  a 

2.  7  +  3:2^  =  37-20;. 

3. 

5(a;- 

3)=a;  +  12. 

4.  6(2; +  5)  = 

10a:. 

5. 

3(8- 

a;)  =  23  +  2a;. 

Solution  of  5. 

24- 

-dx  = 

:  23  +  2a;. 

Transposing  3a;, 

24  = 

:  23  +  2a;  + 

3a; 

=  23 

\  +  5x. 

Transposing  23, 

24- 

23: 

=  62 

•9 

or 

1 

=  5a 

;. 

Dividing  by  5, 

5 

=  x, 

or 

X 

_1 

~5' 

.    Ans. 

6.  7(9  -x)  = 

5(11 

-a;). 

7. 

8(2- 

n)  =  b{n  -  2) 

8.  4(1  -x)  = 

:5(2- 

-X). 

9. 

x-{-a 

=  2a;  +  J. 

10.  3a; -21  + 

4ic- 

7  =  0. 

11. 

ax  —  b  =  0. 

12.  bx  =  c. 

13. 

ax  =  a-{-l. 

14.  {a-\-b)x  = 

:  a  — 

5. 

15. 

mx-\- 

1  =  wa;  +  2. 

16.  A  man  divided  42  apples  between  two  boys,  giving  A 
twice  as  many  as  B.     How  many  did  each  get? 

Solution.     Let  us  call  n  the  number  of  apples  B  got. 
Then,  because  A  got  twice  as  many,  he  got  2n.     So,  by  adding, 
B's  apples  =    n 
A's  apples  =  2n 

Both  together  got  3n 
The  condition  is  that  this  number  shall  be  42.     Hence 
3n  =  42; 
and  dividing  by  3,  ^  =  14. 

Therefore  A's  share  =  2^  =  28; 

B's  share  =    n  =  14=. 

17.  A,  B  and  C  had  between  them  $78.  B  had  twice  as 
much  as  A,  and  C  had  three  times  as  much  as  A.  How  much 
had  each? 

18.  A  drover  bought  a  flock  of  sheep  at  $6  per  head. 
After  losing  40  he  sold  the  remainder  at  $5  per  head  and 
gained  11050.     How  many  did  he  buy? 


70  SIMPLE  EQUATIONS. 

19.  A  man  made  a  journey  of  244  miles  in  3  days,  going 
10  miles  less  on  the  second  day  than  on  the  first,  and  12  miles 
less  on  the  third  day  than  on  the  second.  How  far  did  he  go 
each  day? 

Note,  If  we  call  x  the  distance  made  on  the  first  day,  that  on  the 
second  will  be  a?  —  10  and  that  on  the  third  a?  —  10  —  12  =  «  —  23. 

20.  In  dividing  a  profit  of  $2500  between  two  partners, 
A  got  twice  as  much  as  B  and  1100  more.  What  was  the 
share  of  each? 

21.  A  huckster  had  50  apples,  some  good  and  others  bad. 
He  sold  the  good  at  5  cents  each  and  the  bad  at  2  cents  each, 
realizing  $1. 78  in  all.     How  many  apples  of  each  kind  had  he? 

Method  of  Solution.  Let  us  call  g  the  number  of  good  apples.  Then, 
because  the  good  and  bad  together  numbered  50,  the  bad  alone  num- 
bered 50  —  g.     So  the  sums  realized  from  sales  were: 

g  good  apples  at  5  cents  each 5^ 

50  —  ^  bad  ones  at  2  cents  each 100  —  2^ 

Total  amount  received  ...    100  +  ^9 

This  expression  is  to  be  equated  to  the  amount  realized,  namely,  178 
cents.     By  solving  the  equation  thus  formed,  we  shall  find 
^  —  26  =  number  of  good  apples; 
50  —  ^  =  24  =  number  of  bad  apples. 
26  X  5  4-  24  X  2  =  130  +  48  =  178.     Proof. 

22.  A  man  had  45  pieces  of  coin,  some  3-cent  pieces  and 
the  remainder  10-cent  pieces,  amounting  in  all  to  $1.84.  How 
many  pieces  of  each  kind  were  there? 

23.  A  train  made  a  journey  of  374  miles  in  13  hours,  going 
part  of  the  time  at  the  rate  of  25  miles  an  hour  and  the  re- 
mainder of  the  time  at  the  rate  of  32  miles  an  hour.  How 
many  hours  did  it  run  at  each  rate  of  speed  ? 

Note.  If  we  call  x  the  number  of  hours  it  ran  at  one  rate,  13  —  a; 
will  be  the  number  of  hours  it  ran  at  the  other  rate. 

24.  A  man  made  a  journey  of  112  miles  in  4  days,  going 
4  miles  farther  each  day  than  he  did  the  day  before.  How 
far  did  he  go  on  each  day? 

Note.  Take  the  distance  he  went  on  the  first  day  as  the  unknown 
quantity. 

25.  Divide  100  into  two  parts  such  that  three  times  the 
one  part  shall  be  equal  to  twice  the  other. 


CLEARING  OF  FRACTIONS.  71 

Multiplication  of  Equations. 

86.  Clearing  of  Fractions.  The  operation  of  multipli- 
cation is  usually  performed  in  order  to  clear  the  equation  of 
fractions. 

To  clear  an  equation  of  fractions: 

FiKST  Method.  Multiply  its  memhers  hy  the  least  com- 
mon multiple  of  all  its  denominators. 

Second  Method.  Multiply  its  memhers  hy  each  of  the 
denominators  m  succession. 

Remark  1.  Sometimes  the  one  and  sometimes  the  other  of  these 
methods  is  the  more  convenient. 

Rem.  2.  The  operation  of  clearing  of  fractions  is  similar  to  that  of 
reducing  fractions  to  a  common  denominator. 

Example  of  First  Method.  Clear  of  fractions  the 
equation 

1  +  1  +  1  =  26. 

Here  24  is  the  least  common  multiple  of  the  denominators. 
Multiplying  each  term  by  it,  we  have  (§  68) 

6x-{-^x-{-Sx  =  624, 
or  13a;  =  624. 

Example  op  Secoi^d  Method. 


XX           1 

a'^h~'^' 

Multiplying  by  a, 

.  ax      1 

*  +  T  =  F- 

Multiplying  by  b, 
or 

hx  -]-  ax  =  1, 
(a  4-  h)x  =  1. 

EXERCISES. 

Clear  of  fractions  and  solve  the  following  equations 

1    ^-7  =  -. 
•^•2       ^4 

2.  1  -  2^  =  Sx. 

a:-l_a;  +  l 
^'      2^3- 

^  X      4      a; 

X     '    X     '    X 

X  ^  X     '    X 

72    .                             SIMPLE  EQUATIONS. 
» 

9.   ?4^-+-^=7..  10.  1+1-1=4. 

llJ^^%-^  =  m.  12.  2  .  3;r  -  7  =  0. 
'  a  r   h       c 

^      li\  apx  -  ^^  =  0.  14.  1  _  ?^  r.  3. 

/                                               .  X       mx 

15. r— =  3.  lo.  — ^ -. =  d.4. 

<      17.-^^ j-=«J.        18.^ 1  =  ^  +  1. 

19.  «-^^~^  =^y  +  7.  30.  4k +  13.'!;  =  4.  12. 

o  7 


Problems  Leading  to  Simple  Equations. 

To  solve  a  problem,  we  hnYe  first  to  state  the  conditions  of 
the  problem  in  algebraic  language,  and  secoyid  to  solve  the 
equation  resulting  from  such  statement. 

We  have  already  shown  how  to  solve  the  equation,  but  for 
the  statement  no  general  rule  can  be  laid  down.  The  follow- 
ing precepts  will,  however,  serve  as  a  guide  to  the  beginner, 
who  must  trust  to  practice  to  acquire  skill  in  solving  problems. 

1.  Study  the  problem  carefully  to  see  what  is  the  unknown 
quantity  required  to  be  found.  Sometimes  there  are  several, 
only  one  of  which  need  be  taken. 

2.  Represent  this  quantity  by  x,  y,  or  any  other  letter  or 
symbol  whatever.* 

3.  Perform  on  and  with  this  symbol  the  operations  de- 
scribed in  the  problem  (as  in  Chapter  I.,  §  23). 

4.  Express  the  conditions,  stated  or  implied,  in  the  prob- 
lem by  means  of  an  equation. 

5.  Solve  the  equation  by  the  methods  already  explained  in 
the  last  four  sections. 

*  Any  symbol  may  be  used.  In  the  early  history  of  algebra  the  un- 
known quantity  was  called  the  thing — in  Italian  cosa,  which  for  brevity 
was  written  co. 


UNIVERSITY 

OP 

CAffPg^^a^  LEADING   TO  SIMPLE  EQUATIONS.       73 


I  *• 


PROBLEMS    FOR    PRACTICE. 

1.  A  man  asked  a  shepherd  how  many  sheep  he  had.  He 
replied :  If  you  add  32  to  the  number  of  my  sheep  and  multi- 
ply the  sum  by  9,  the  product  will  be  27  times  the  number 
of  my  sheep. 

2.  Another  shepherd  answered:  If  you  subtract  32  from 
my  sheep  and  multiply  the  remainder  by  6,  you  will  have 
double  the  number. 

3.  A  baker  said:  If  you  divide  the  number  of  my  loave$ 
by  6  and  add  60  to  the  quotient,  the  sum  will  be  the  number^' 
of  my  loaves. 

4.  A  man  sets  out  upon  a  journey  from  one  city  to  an- 
other. The  "first  day  he  travels  one  half  the  distance  between 
the  cities;  the  next  day  he  travels  one  third  the  distance  be- 
tween the  cities,  and  then  finds  he  has  still  12  miles  to  go. 
Find  the  distance  between  the  cities. 

Solution,     Let  d  =  the  distance  between  the  cities;  then 

-  =  the  distance  travelled  the  first  day, 

-  =    **        "  "         "  second  day, 
o 

and  these  two  amounts  plus  12  miles  must,  by  the  conditions 
of  the  problem,  equal  the  whole  distance  d;  that  is,  in  alge- 
braic language  the  conditions  of  the  problem  are  stated  in 
the  following  equation: 

f  +  f  +  12  =  .. 

To  solve  this  equation  multiply  each  member  by  6,  whence 
3c?  +  2g?  +  72  =  6d. 
Transposing,  72  =  Gc?  —  3^7  —  2d; 

and  uniting  terms,  72  =  d. 

Hence  72  miles  is  the  required  distance  between  the  cities. 

72 
Froof.  Distance  travelled  the  first  day  =  --  =  36  miles; 

72 
"  "        second   *'    =  —  =  24  miles; 

"  "        third      ''  =  12  miles. 

Whole  distance  travelled  =  72  miles. 


4 


74  SIMPLE  EQUATIONS. 

5.  Another  said:  If  you  add  9  to  three  times  the  number 
of  my  loaves  and  divide  the  sum  by  9,  the  quotient  will  be  38. 

6.  A  third  baker  said:  If  you  add  136  to  twice  the  number 
of  my  loaves  and  divide  the  sum  by  the  number  of  my  loaves, 
the  quotient  will  be  10. 

7.  If  you  add  41  to  a  number  and  divide  the  sum  by  the 
number  minus  15,  the  quotient  will  be  8.  What  is  the  mim- 
ber?  Ans.  23. 

8.  If  you  add  96  to  6  times  a  number  and  divide  the  sum 
by  twice  the  number  minus  52,  the  quotient  will  be  17.  Find 
the  number.  Ans.  35. 

9.  If  you  subtract  3  times  a  certain  number  from  480  and 
divide  the  difference  by  the  number  minus  18,  the  quotient 
will  be  139.     Find  the  number.  'Ans.  21. 

10.  Find  a  number  such  that  if  we  divide  it  by  6,  add  7 
to  the  quotient,  and  multiply  the  sum  by  two,  the  product 
will  be  the  number  itself.  Ans.  21. 

11.  From  the  following  condition  find  the  age  at  which  Sir 
Isaac  Newton  died:  If  you  take  one  third  his  age,  one  fourth 
his  age,  and  one  sixth  his  age,  then  add  them  together,  and 
add  105  to  the  sum,  you  will  get  just  double  his  age.     Ans.  84. 

12.  If  you  add  5  to  the  year  in  which  Sir  Isaac  Newton 
was  born,  then  divide  the  sum  by  9  and  add  638  to  the  quo- 
tient, the  sum  will  be  just  half  the  number  of  the  year. 
What  was  the  year?  Ans.  1642. 

13.  If  you  take  the  year  in  which  John  Hancock  died, 

add  16  to  it,  then  take  ^,  -,  and  ^  of  the  sum,  the  sum  of 

these  three  quotients  will  be  871.   Find  the  year.  Ans.  1793. 

14.  If  you  subtract  the  age  of  John  Hancock  from  100, 
divide  the  difference  by  4,  and  add  the  quotient  to  one  half 
his  age,  the  sum  will  be  17  years  less  than  his  age.  Find  liis 
age.  Ans.  56. 

15.  If  you  take  the  population  of  the  District  of  Colum- 
bia in  1870,  divide  it  by  300,  subtract  400  from  the  quotient, 
and  multiply  the  remainder  by  7,  the  product  will  be  273. 
What  was  the  population?  Ans.  131,700. 

16.  If  you  divide  my  age  10  years  hence  by  my  age  21 
years  ago,  the  quotient  will  be  2.     What  is  my  present  age? 


PBOBLEMS  LEAUij-sG   TO  SIMPLE  EQUATIONS.      75 

17.  Divide  1200  among  three  persons,  A,  B  and  C,  so 
that  B  shall  have  $25  more  than  A,  and  C  $18  less  than  A 
and  B  together. 

Solution.  This  question  differs  from  tliose  preceding  in  having 
three  known  quantities.  We  therefore  show  how  to  solve  it.  Let  us  put 
y  for  A's  share.  Then  B's  share  will  hey-\-  25,  and  the  share  of  A  and  B 
together  will  be  2/  +  y  +  35;  that  is,  2y -\-  25.  C's  share  is  said  to  be 
$18  less  than  this  sum;  it  is  therefore  2y -f  7.     We  now  add  up  the 

shares : 

A's  share  is   y 

B's     -     -    y  +  25 

C's     "     "  2y+    7 

Sum  of  all,  4y  +  32 

Kow  by  the  conditions  of  the  problem  this  sum  must  be  $200. 

Hence  we  put  it  equal  to  200  and  solve  the  equation.     We  shall  thus 

find  y  =  42.     Therefore  we  have  for  the  answer: 

A's  share  =     y  =42 

B's     "     =     y  -f  25  =    67 

C's      •'      =  109  -  18  =    91 

Total,  $200.    Proof.  / 

18.  A  father  left  $7050  to  be  divided  among  five  children, 
directing  that  the  eldest  should  have  $200  more  than  the 
second,  the  second  $200  more  than  the  third,  and  so  on  to  the 
youngest.     What  was  the  share  of  each? 

Ans.  $1810,  $1610,  $1410,  $1210,  $1010. 

19.  A  is  15  years  older  than  B,  and  in  18  years  A  will  be 
just  twice  as  old  as  B  is  now.     What  are  their  ages? 

Ans.  48,  33. 

20.  Of  three  brothers  the  youngest  is  4  years  younger  than 
the  second,  and  the  eldest  is  as  old  as  the  other  two  together. 
In  10  years  from  now  the  sum  of  their  ages  would  be  98. 
What  are  their  ages  now?  Ans.  15,  19,  34. 

21.  An  uncle  left  his  property  to  two  nephews,  the  elder 
to  receive  $100  more  than  the  younger.  But  if  a  certain 
cousin  was  still  living,  the  elder  was  to  give  her  one  fourth 
of  his  share,  and  the  younger  one  sixth  of  his.  The  cousin 
was  living,  and  got  $1275.  How  much  did  the  uncle  leave, 
and  what  was  the  share  of  each  heir?      Ans.  $3000,  $3100. 

22.  The  head  of  a  fish  is  9  inches  long,  the  tail  is  as  long 
as  the  head  and  half  the  body,  and  the  body  is  as  long  as  the 


76  SIMPLE  EQUATIONS. 

head  and  tail  together.      What  is  the  whole  length  of  the 
fish?  Ans.  6  feet. 

23.  Divide  the  number  104  into  two  such  parts  that  one 
eighth  of  the  greater  part  shall  be  equal  to  one  fifth  of  the 
lesser.  Ans.  64,  40. 

Note.     If  one  part  be  x,  the  other  will  be  104  —  x. 

24.  Divide  188  into  two  such  parts  that  the  fourth  of  one 
part  may  exceed  the  eighth  of  the  other  by  18. 

25.  Two  gamblers,  A  and  B,  engage  in  play.  When  they 
begin  A  has  $200  and  B  has  $152.  When  they  finish  A  has 
three  times  as  much  as  B.     How  much  did  he  win? 

26.  A  father  has  five  sons  each  of  whom  is  5  years  older 
than  his  next  yo anger  brother,  and  the  eldest  is  five  times  as 
old  as  the  youngest.     What  are  their  ages? 

27.  A  father  is  now  4  times  as  old  as  his  son,  but  in  four 
years  he  will  only  be  3  times  as  old.     How  old  is  each? 

28.  The  sum  of  $345  was  raised  by  A,  B  and  C  together. 
B  contributed  twice  as  much  as  A  and  $15  more,  and  0  three 
times  as  much  as  B  and  $15  more.  How  much  did  each 
raise? 

29.  Divide  the  number  97  into  two  parts  such  that  twice 
the  one  part  added  to  three  times  the  other  shall  be  254. 

Ans.  37,  60. 

30.  A  father  divided  his  estate  among  four  sons,  directing 
that  the  youngest  should  receive  -J-  of  the  whole;  the  next, 
$1000  more;  the  next,  as  much  as  these  two  together;  and 
the  eldest,  what  was  left.  The  share  of  the  eldest  was  $5000. 
What  was  the  value  of  the  estate,  and  the  shares  of  the  three 
younger?  Ans.  $14,000;  $1750,  $2750,  $4500. 

31.  An  almoner  divided  $86  among  59  people,  giving  the 
barefoot  ones  $1  each  and  the  others  $1. 75  each.  How  many 
barefoot  ones  were  there? 

32.  An  almoner  divided  $75  among  a  crowd  of  people, 
giving  one  third  of  them  50  cents  each,  one  fourth  $1  each, 
and  the  remainder  $1.50  each.  How  many  people  were 
there? 

33.  A  father,  in  making  his  will,  directs  that  the  second 
of  his  three  sons  shall  have  half  as  much  again  as  the  young- 


PROBLEMS  LBADINa  TO  SIMPLE  EQUATIONS.      77 

est,  and  the  eldest  one  third  more  than  the  second.  The 
eldest  receives  $2250  more  than  the  youngest.  What  was  the 
share  of  each? 

34.  Divide  the  number  144  into  four  parts  such  that  the 
first  part  divided  by  3,  the  second  multiplied  by  3,  the  third 
diminished  by  3,  and  the  fourth  increased  by  3  shall  all  four 
be  equal  to  each  other. 

Call  X  tlie  quantity  to  which  these  four  results  are  equal.  Then  the 
part  which  divided  by  3  will  make  x  must  be  ^x,  the  second  part  must 

X 

be  =,  the  third  x-\-Z,  and  the  fourth  a?  —  8.     To  form  the  equation  note 

o 

the  sum  of  all  the  parts  is  144,  which  gives  the  equation  to  be  solved. 

Ans.  81,  9,  30,  24. 

35.  Divide  the  number  100  into  four  parts  such  that  the 
first  being  multiplied  by  4,  the  second  divided  by  4,  the  third 
increased  by  4,  and  the  fourth  diminished  by  4  shall  give 
equal  results. 

36.  A  man  making  a  journey  went  on  the  first  day  one 
third  of  the  distance  and  36  miles  more;  on  the  second,  one 
third  the  remaining  distance  and  36  miles  more,  which  jast 
brought  him  to  his  journey's  end.  What  was  the  length  of 
the  journey? 

37.  What  number  of  apples  was  divided  among  three 
people  when  the  first  was  given  half  the  apples  and  half  an 
apple  more,  the  second  half  of  what  remained  and  half  an 
apple  more,  and  the  third  half  of  what  then  remained  and 
half  an  apple  more,  which  emptied  the  basket? 

38.  In  the  division  of  an  estate  between  A,  B,  C  and  D, 
A  got  one  tenth  of  the  whole,  B  got  half  as  much  as  A  and 
$2505  more,  0  got  half  as  much  as  B  and  $2505  more,  and 
D  got  $4050  dollars.     What  was  the  amount  of  the  estate? 

39.  A  person  journeying  to  a  distant  town  travelled  on 
the  first  day  half  way,  wanting  20  miles;  on  the  second  day 
half  the  remaining  distance,  wanting  5  miles;  and  on  the 
third  day  half  the  distance  still  remaining  and  10  miles 
more,  when  he  still  had  12  miles  to  travel  on  the  fourth  day. 
How  long  was  his  whole  journey? 


78  SIMPLE  EQUATIONS. 

40.  A  trader  having  made  a  profit  of  r  per  cent  on  his 
capital  found  that  capital  and  profits  together  amounted  to  a 
dollars.     What  was  his  capital? 

Note.  By  the  definition  of  percentage,  r  per  cent  of  a  sum  means 
r  hundredths  of  that  sum.     Hence  r  per  cent  is  found  by  multiplying 

by  T^.     If  the  capital  is  c,  r  per  cent  of  it  is  :7^,  and  when  this  is  add- 
ed  to  the  capital  the  sum  will  be  6  +  r^. 

41.  A  trader  having  increased  his  capital  by  8  per  cent 
found  that  it  amounted  to  $4320.     How  much  was  it  at  first? 

42.  A  merchant  increased  his  capital  by  8  per  cent  the 
first  year  and  increased  that  increased  capital  by  12  per  cent 
the  second  year,  when  the  total  amounted  to  m  dollars.  How 
much  had  he  at  first? 

43.  Of  three  casks  the  second  contained  15  per  cent  more 
than  the  first,  and  the  third  20  per  cent  more  than  the  second. 
The  first  and  third  together  contained  119  gallons.  How 
much  did  the  second  contain? 

44.  A  man  investing  a  sum  of  money  got  5  per  cent  inte- 
rest on  it  the  first  year,  which  he  added  to  his  principal,  and 
then  got  4  per  cent  on  the  amount  for  the  second  year.  At 
the  end  of  the  second  year  he  found  interest  for  the  two  years 
to  amount  to  $2457.     How  much  did  he  invest? 

45.  Of  two  men  A  and  B,  A  had  in  money  25  per  cent 
more  than  B.  B  having  received  $225  then  had  25  per  cent 
more  than  A.     How  much  had  each  at  first? 

46.  The  perimeter  of  a  triangle  measures  320  yards.  The 
first  side  is  40  yards  longer  than  the  base,  and  the  second  side 
is  half  the  sum  of  the  base  and  first  side.  What  is  the  length 
of  each  side? 

Note,  By  the  perimeter  of  a  triangle  is  meant  the  sum  of  its  three 
sides.    The  base  is  any  one  of  the  three  sides. 

47.  The  perimeter  of  a  triangle  measures  'p  feet.  The  first 
side  is  m  feet  longer  than  the  base,  and  the  remaining  side  is 
m  feet  longer  than  the  first  side.  What  are  the  lengths  of  the 
three  sides? 

48.  A  pedestrian  made  a  journey  of  125  miles  in  5  days, 
going  3  miles  less  on  each  day  than  he  did  on  the  day  preced- 


PROBLEMS  LEADING   TO  SIMPLE  EQUATIONS      79 

ing.  How  far  did  lie  go  on  the  first  day?  How  far  on  the 
last  day? 

49.  A  man  bought  3  works  each  containing  a  volumes, 
and  2  works  each  containing  h  yolumes,  for  x  dollars.  What 
was  the  price  of  each  volume? 

50.  A  man  bought  a  work  of  m  volumes  for  x  dollars.  How 
much  would  h  volumes  cost  at  the  same  price  per  volume? 

51.  A  line  64  feet  long  is  divided  into  three  parts  such 
that  the  second  part  is  one  fourth  longer  than  the  first,  and 
the  third  part  one  fourth  shorter  than  the  first  and  second 
together.     What  is  the  length  of  each  part? 

52.  A  factory  employed  men  at  $2  per  day,  38  more  women 
than  men,  paying  them  $1  per  day,  and  35  more  boys  than 
women,  paying  the  boys  60  cents  per  day.  The  daily  wages 
amounted  to  $197.  How  many  operatives  were  there  of  each 
class? 

53.  In  another  factory  there  were  one  third  more  women 
than  men,  and  one  third  more  boys  than  men  and  women 
together.  The  wages  were:  men  $1.50,  women  $1,  boys  50 
cents, — making  a  sum  total  of  $276.50.  How  many  operatives 
were  there  of  each  class? 

54.  Four  men  having  received  a  sum  of  D  dollars  to  be 

divided  between  them,  the  first  2:ot  —  of  the  whole,  and  the 

second  half  as  much  as  the  first  and  x  dollars  more;  the 
remainder  being  divided  equally  between  the  third  and  fourth. 
How  much  did  each  get? 

55.  If  from  a  sum  of  x  dollars  I  take  one  third,  then  one 
half  of  what  is  left,  and  then  one  third  of  what  is  still  left, 
how  much  remains? 

56.  Having  an  equation 

ax  =  5, 
I  want  to  put  for  a  and  h  two  numbers  whose  sum  shall  be 
20,  and  which  shall  give  f  for  the  value  of  x.    What  numbers 
shall  I  use  for  a  and  h? 

57.  In  another  equation  of  the  same  form  I  want  h  to  be 
greater  than  a  by  12,  and  the  value  of  x  to  be  2.  What  num- 
bers shall  I  use  for  a  and  J? 


80 


SIMPLE  EQUATIONS. 


58.  In  the  fraction 


m-\-x 


dm  +  X 
I  want  to  put  for  x  such  a  quantity  that  the  value  of  the 
fraction  shall  be  f.     What  value  of  x  shall  I  use? 

59.  What  must  be  the  value  of  m  in  order  that  the  sum  of 

the  two  fractions  — -—  and  — --^  may  be  —  ? 
m  + 1         m-{-l       ''■       /J 

60.  Out  of  a  bin  of  wheat  one  man  took  one  fourth  of  the 
wheat  and  3  bushels  more,  and  a  second  took  one  third  of 
what  was  left  and  5  bushels  more.  What  he  left  was  5 
bushels  more  than  the  first  man  took.  How  much  wheat  was 
in  the  bin,  and  how  much  did  each  take? 


Memoranda  aitd  Exercises  for  Eeyiew. 

Define :  Equation;  Identical  equation;  Equation  of  con- 
dition ;  Solving  an  equation ;  Root;  Axiom. 

Axioms  of  addition  and  subtraction. 
Rule  for  transposition;  G-ive  examples. 
Prove  the  general  rule. 
Write  an  equation  which  may  be  solved  by 

simple  transposition  and  aggregation  of 

terms. 


Addition 
and  Sub- 
traction. 


Division. 


Multipli- 
cation. 


Axiom  of  division. 

Solution  of  an  equation  by  division;  Rule. 
Write  an  equation  with  four  terms  which 
can  be  solved  by  transposition  and  division. 

C  Axiom  of  multiplication. 
^  When  multiplication  is  required. 
1^  Two  methods;  Explain  both. 


SEOOKD    OOUESE. 


ALGEBEA   TO    QTJADRATIO 
EQUATIONS. 


CHAPTER  I. 
THtORY  OF  ALGEBRAIC  SIGNS. 


Use  of  Positive  and  Negative  Signs. 

88.  Opposite  Directions  of  Measurement.  Most 
of  the  quantities  we  have  to  express  in  algebraic  lan- 
guage may  be  measured  in  two  opposite  directions. 

Examples.     Ti7ne  may  be  either  time  before  or  time  after. 

Distance  on  a  straight  road  may  be  measured  from  any 
point  in  two  opposite  directions.  If  the  road  is  east  and 
west,  one  direction  will  be  west  and  the  other  east. 

The  scale  of  a  thermometer  is  divided  so  as  to  measure 
from  zero  in  two  opposite  directions. 

89.  Positive  and  Negative  Measures.  In  order 
to  measure  such  quantities  on  a  uniform  system,  the 
numbers  of  algebra  are  considered  to  increase  from  0 
in  two  opposite  directions.  Those  in  one  direction  are 
called  positive;  those  in  the  other  direction,  nega- 
tive. 

Positive  numbers  are  distinguished  by  the  sign  +, 
negative  ones  by  the  sign  — . 

If  a  positive  number  measures  years  after  Christ,  a  nega- 
tive one  will  mean  years  before  Christ. 

If  a  positive  number  is  used  to  measure  toward  the  right, 
a  negative  one  will  measure  toward  the  left. 

If  a  positive  number  measures  weight,  the  negative  one 
will  imply  lightness,  or  tendency  to  rise  from  the  earth. 

If  a  positive  number  measures  property,  or  credit,  the 
negative  one  will  imply  debt. 


84  THEORY  OF  ALGEBRAIC  SIGNS. 

90.  The  series  of  algebraic  numbers  will  therefore 
be  considered  as  arranged  in  the  following  way,  the 
series  going  out  to  infinity  in  both  directions : 


.^   Negative  Sirootioa. 

Positive  Bireotion.    ^ 

Before. 

After. 

Downward. 

Upward. 

Debt. 

Credit. 

etc. 

etc. 

5,  -4,  -3,  -2,  - 

-1, 

0, 

+1, 

,  +2,  +3,  +4,  +5,  etc. 

etc. 

91.  Clioice  of  Direction.  It  matters  not  which 
direction  we  take  as  the  positive  one,  so  long  as  we 
take  the  opposite  one  as  negative. 

If  we  take  time  lefore  as  positive,  time  after  will  be  nega- 
tive; if  we  take  west  as  the  positive  direction,  east  will  be 
the  negative;  if  we  take  debt  as  positive,  credit  will  be  nega- 
tive. 

93.  The  Scale  of  Numbers.  Positive  and  nega- 
tive numbers  may  be  conceived  to  njeasure  distances 
from  a  fixed  point  on  a  straight  line,  extending  in- 
definitely in  both  directions,  the  distances  on  one  side 
being  positive,  on  the  other  side  negative,  as  in  the 
following  diagram :  * 

etc. -7,  -6,  -5,  -4,  -3.  -2,  -1,    0,-f  1,  +2,  +3,  -|-4, 4-5. -j-6.  +7, etc. 

I — i — I — I — \ — \ — ! — [— i — \ — I — \ — I — \ — I 

Positive  direction. 

■>■ >. 


Negative  direction. 

< 


In  the  above  scale  of  numbers  the  positive  direc- 
tion is  from  left  to  right  and  the  negative  direction 
from  right  to  left^  as  shown  by  the  arrows ;  and  the. 
distance  between  any  two  consecutive  numbers  is  con- 
sidered a  unit  or  unit  step. 

*  The  student  should  copy  this  scale  of  numbers,  and  have  it  before 
him  in  studying  the  present  chapter. 


MEANING   OF  MINUS.  85 

93.  Def.  In  the  scale  of  numbers  any  number 
which  lies  in  a  positive  direction  from  another  is  called 
algebraically  greater  than  that  other.     Thus, 

—  2    is  algebraically  greater  than     —  7  ; 

0     "  "  "  "        -2', 

5     "  "  "  "        -5. 

Def,  The  absolute  value  of  a  quantity  is  its  value 
without  regard  to  its  algebraic  sign. 

Example.  The  absolute  vahie  of  —  3  is  the  same  as  the 
absolute  value  of  +  3;  namely,  3  without  any  sign. 

94.  Meaning  of  Minus.  We  now  have  the  following 
new  definition  of  the  minus  sign: 

Minus  means  opposite.  A  minus  sign  shows  that 
the  quantity  before  which  it  is  placed  must  be  taken 
in  the  opposite  sense  from  that  in  which  it  would  be 
taken  if  the  sign  were  not  there. 

Example  1.   —  x  is  the  opposite  of  x. 

Ex.  2.    —  (—  a;)  is  the  opposite  of  —  x;  that  is,  it  is  x. 
Therefore  —  {—  x)  =  x. 

Ex.  3.  —  [—  (—  ^)]  is  the  opposite  of  —  (—  x),  or  of  x. 
Hence  —  [—  {—  x)]  =  —  x.     etc.  etc. 

EXERCISES. 

What  is  the  meaning  of: 

1.  —  n  years  after  Christ? 

2.  —  (—n)  years  after  Christ? 
S.   —n  years  before  Christ? 

4.  —  {—  n)  years  before  Christ? 

5.  The  thermometer  is  —  15°  above  zero? 

6.  The  thermometer  has  risen  —  20°? 

7.  The  thermometer  has  fallen  —  (—  20°)? 

8.  To-day  is  —  25°  colder  than  yesterday?     ^ 

9.  To-day  is  —  25°  warmer  than  yesterday? 

10.  To-day  is  —  (—  25°)  warmer  than  yesterday? 

11.  John  lives  —  2  miles  east  of  William? 

Ans.     John  lives  2  miles  west  of  William. 

12.  James  lives  —  4  miles  south  of  John? 


86  THEORY  OF  ALGEBRAIC  SIGNS. 

13.  Smith  is  —  6  years  older  than  Jones? 

14.  Jones  is  —  5  years  younger  than  Brown? 

15.  John  owes  the  grocer  —  $5? 

16.  Thomas  weighs  —  12  pounds  more  than  Jane? 

17.  Mr.  Weston  is  —  $1000  richer  than  Mr.  Brown? 
Answer  the  following  questions  in  algebraic  language: 

18.  James  owed  William  $12  and  paid  him  $7.  How 
much  did  he  still  owe?     How  much  did  William  owe  James? 

19.  If  James  had  owed  $12  and  paid  $15,  how  much  would 
he  still  owe? 

20.  If  he  owed  William  a  dollars  and  paid  him  b  dollars, 
how  much  would  he  still  owe?  How  much  would  William 
owe  him? 

21.  The  thermometer  was  15°  on  Sunday,  and  next  day  it 
was  22°  lower.     What  was  it  then? 

22.  What  day  is  the  0th  of  February?  The  -  1st?  The 
-4th?    The -31st? 

Note  that  the  0th  day  is  the  day  before  the  1st  day. 

95.  Def.  When  two  quantities  are  numerically 
equal  but  with  opposite  signs,  each  is  said  to  be  the 
negative  of  the  other. 

Examples.      —  a  is  the  negatiye  of  a. 

a  is  the  negative  of  —  a. 

X  —  y  is  the  negatiye  oi  y  —  x, 

EXERCISES. 

What  is  the  negative  of: 

1.  3a?  2.   —25?  3.  x—a  —  h'^ 

4.   -2«/  +  3a-5?  5.   -(«-5)? 

96.  Lines  to  represent  Nwrnbers.  A  number  may 
be  represented  to  the  eye  by  drawing  a  line  long 
enough  to  reach  from  the  zero  point  to  the  number  on 
any  fixed  scale. 

To  fix  the  scale  a  length  must  be  assumed  as  a 
unit,  and  a  direction  must  be  chosen  as  positive.  The 
line  must  then  have  as  many  units  as  the  number  to 
be  represented. 

A  negative  number  is  drawn  in  the  opposite  direc- 
tion from  a  positive  one. 


ALGEBIIAIC  ADDITION.  87 

Examples.     If  we  take  this  length  as  the  unit, 

-f  3  is  this  line  o 

—  2  this  hne o 

and  —  1  this  line  o 

The  zero  point  is  always  the  beginning  of  the  line. 

V  ♦  EXERCISES.  < 

Taking  any  unit  you  please,  draw,  by  the  eye,  lines  to 
represent: 

1.  +2.         2.   +5.         3.   -4.         4.    -3.         5.    -1. 

Algebraic  Addition. 

97.  Algebraic  addition  is  the  operation  of  com- 
bining quantities  according  to  their  algebraic  signs, 
and  is  performed  by  taking  the  difference  between 
the  sums  of  the  positive  and  negative  quantities  and 
prefixing  the  sign  of  the  greater  sum. 

Def.  The  result  of  algebraic  addition  is  called  the 
algebraic  sum. 

Example.     The  algebraic  sum  of  4  -j-  5  —  7—  8  +  2  is 
+  11  -15  =  -4. 

98.  Algebraic  addition  is  represented  on  the  scale 
of  numbers  by  measuring  off  the  positive  quantities 
in  the  positive  direction,  and  the  negative  quantities 
in  the  opposite  direction,  commencing  each  measure 
at  the  end  of  the  one  preceding. 

Example.     To  represent  the  algebraic  sum 
4-6  +  3-5-1, 
we  start  from  0  and  measure  off  4  unit  steps  to  the  right, 
which  takes  us  to  +  4.    Then  we  measure  6  from  +  4  toward 
the  left,  which  brings  us  to  —  2.     Then  3  to  the  right  brings 
us  to  + 1.     Then  5  to  the  left  brings  us  to  —  4.     Then  1  to 
the  left  brings  us  to  —  5,  which  is  the  algebraic  sum.     Thus 
4-6  +  3-5-1  =  - 5. 

99.  Formation  of  Algebraic  Sums.  We  see  from  the 
above   that   the  algebraic  sum  of  several  numbers  may  be 


88  THEORY  OF  ALGEBRAIC  SIGNS. 

formed  by  putting  the  lines  which  represent  them  end  to  end 
in  their  proper  directions,  the  beginning,  or  zero  point,  of  each 
one  being  at  the  end  of  the  preceding  one. 

The  preceding  example    (+  4  —  6  -f  3  —  5  —  1)   is  then 
represented  in  this  way: 

Sum  =  o '■ 


Examples  of  Algebraic  Sums.  Three  traders  agree  to 
divide  their  profits  and  losses  equally.  The  first  year  A 
gained  13000,  B  $4000  and  C  $8000.  The  algebraic  sum  of 
the  profits  to  be  divided  was  therefore 

$3000  +  14000  +  18000  =:  $15,000. 
Each  man's  share  =  $5000. 

Next  year  A  gained  $4000  and  B  $5000,  while  C  lost  $3000. 
The  algebraic  sum  to  be  divided  was  therefore 
$4000  +  $5000  -  $3000  =  $6000. 
Each  man's  share  =  +  $2000. 

The  third  year  A  gained  $2000,  B  lost  $1000  and  C  lost 
$4000.     The  algebraic  sum  of  their  profits  was  therefore 

$2000  -  $1000  -  $4000  ==  -  $3000. 
Therefore  each  man  got  —  $1000;  that  is,  he  suffered  a  loss 
of  $1000. 

EXERCISES. 

Draw  a  scale  of  numbers  (§  92)  from  —  8  to  -j-  8,  and  then 
form  the  following  algebraic  sums  by  lines  under  the  scale: 
1.-4-^6-3+5  +  1.  2.        1-2  +  3-4. 

3.       5-3-3-3-3.  4.-4-4  +  3  +  4  +  5. 

5.  The  temperature  at  9  points  scattered  equally  over  the 
country  is  +15°,  +20%  -8°,  -13°,  -2°,  +7°,  -15°, 
—  10°  and  —  12°.     What  is  the  mean  temperature? 

Note.  The  mean  of  any  series  of  numbers  is  obtained  by  dividing 
their  algebraic  sum  by  the  number  of  terms. 

6.  A  merchant  in  5  successive  years  lost  $3000,  made 
$2000,  made  $8000,  lost  $1000  and  made  $3000,  What  was 
bis  average  aiiiiuul  ])ryfit? 


ALGEBRAIC  SUBTRACTION.  89 

Algebraic  Subtraction. 

100.  Subtraction  in  algebra  consists  in  expressing 
tbe  algebraic  difference  between  two  quantities. 

Def.  The  algebraic  difference  of  two  quantities 
is  their  number  of  units  ajjart  on  the  scale  of  numbers. 

EXERCISES. 

What  is  the  algebraic  difference  between: 
1.-3  and  +5?        Ans.     8. 
2.   +3  and +5?  3.   -7  and -3? 

4.   +  5  and  -  3?  5.-9  and  +  9? 

101.  Sign  of  Remainder.  Tlie  sign  of  the  alge- 
braic difference  is  shown  by  the  direction  from  the 
subtrahend  to  the  minuend. 

Examples.    From  —8  to  — 3is+c-.  —  3  —  (—  8)  =  +5. 
From  -3  to  -8  k  ~.'.-~S  —  {-3)  =  -b. 

Remark,  When  we  subtract  a  lesser  positive  quantity  from  a 
greater  one,  this  rule  gives  a  positive  remainder,  so  that  the  result  is 
the  same  as  in  arithmetic.  But  the  algebraic  process  takes  account  of 
cases  whicli  arithmetic  does  not;  for  example,  tliose  where  the  subtra- 
hend is  greater  than  the  minuend.  This  reversal  of  the  case  is  indi- 
cated by  the  negative  sign  of  the  difference. 

EXERCISES. 

Give  the  values  of  the  three  quantities  a-\-If,  a  —b  and 
b  —  a  in  the  following  cases: 

1.  When  a  =  +  13  and  h  =  -{-^. 

2.  When  a  =  -f  13  and  ^  =  —  8. 

3.  When  a  =  -  13  and  ^  =  -f-  8. 

4.  When  a  =  -  13  and  b  =  ~  8. 

5.  Of  two  bankrupts,  A  and  B,  A  owes  15000  more  than 
all  he  possesses,  and  B  owes  $3000  more.  Which  is  the 
richer,  and.  how  much  richer  is  he? 

6.  In  this  case,  what  is  the  sum  of  their  possessions? 

7.  Of  two  merchants,  A  and  B,  A  made  $2000  and  then 
lost  $3000;  B  lost  $2000  and  then  made  $7000.  How  much 
more  was  A  worth  than  B?  How  much  more  was  B  worth 
than  A? 


90 


THEORY  OF  ALGEBRAIC  SIGNS. 


8.  During  two  years  A  gained  a  dollars  and  afterward  lost 
b  dollars;  B  first  gained  x  dollars  and  then  lost  y  dollars. 
Express  how  much  A  was  worth  more  than  B. 

Rule  of  Signs  in  Multiplication. 

102.  I.  In  multiplying  a  line  by  a  positive  factor,  we 
leave  its  direction  unchanged. 

II.  In  multiplying  by  a  negative  factor,  we  change  the 
direction  of  the  line. 

Illustration.  Suppose  the  quantity  a  to  represent  a  length 
of  one  centimetre  from  the  zero  point  toward  the  right  on 
the  scale  of  §  92.     Then  we  shall  have 

0 

a  =  this  line    i  i 

The  products  of  this  line  by  the  factors  from  -]-  2  to  —  3 

will  be: 

0 

rt  X  2 I  I  I  =  +  2a. 


a  X  1  . 

a  X  0  . 

a  X  -  1 

a  X  —  2 

a  X  -  3 


zzz 

a. 

= 

0. 

= 

— 

a. 

= 

— 

2a, 

3:: 



3a. 

We  shall  also  have 

0 

—  a  =  this  line    |  | 

The  products  of  this  line  by  the  factors  from  -f-  3  to  —  2 
will  be: 

—  a  X3     .     . 

—  a  X  2    .     . 

—  a  X  1     .     . 

—  a  X  0     .     . 

—  a  X  -  1     . 

—  a  X  -  2     . 


= 

-  3a. 

= 

-  2a. 

= 

—  a. 

= 

0. 

— 

+  «. 

= 

+  2a. 

The  preceding  result  may  be  expressed  thus: 
In  muUipUcation  like  siffus  give  plus. 


IIJILE  OF  SIGNS  IN  DIVISION.  91 

103.  Wlien  there  are  tliree  or  more  negative  factors,  the 
product  of  two  of  them  will  be  -f^  the  third  will  change  the 
sign  to  — ,  the  fourth  will  change  this  —  to  -|-  again,  and  so 
on.     Hence: 

The  proihict  of  an  evei^  number  of  negative  factors  is  posi- 
tive. 

The  product  of  an  odd  number  of  negative  factors  is  nega- 
tive, 

EXERCISES. 

Form  the  following  products  by  the  method  of  §  37,  and 

assign  the  proper  signs: 

1.    -  2  X  3  X  -  4.  '     2.   3  X  -  5  X  -  G  X  -  «. 

3.   —  a  X  —  a'^x  X  —  a^x^.  4.  m  X  cm''  X  —  c^x. 

5.    -  2Z'  X  3Z*  X  ^bc.  6.   3^  X  -  hab  X  Ic 

7.   -  5  X  -  6  X  -  7  X  -  8.  8.   -  1  x  -  2  X  -  3  X  «^ 

9.    -lX-lX-1.  10.    -lX-lX-lX-1. 

11.    —  ahc  X  —  ah  X  —aXx,  12.  m  X  —  1  X  m  X  nf. 

13.  (-l)^       14.  {-ly.     15.  (-ly.       le.  {-ly. 

17.   (-l)^  18.  --X-'' 


n  m 


a 


b'  .      a 


1 0.    ^-  X  -  -^3  X  -  ab  X  y         20.   (-  1)^^ 


Rule  of  Sigfiis  in  Division. 

104.  The  rule  of  signs  in  division  corresponds  to  that 
in  multiplication,  namely: 

If  dividend  and  divisor  have  the  same  sign,  the  quotient 
is  positive. 

If  they  have  opposite  signs^^  the  quotient  is  negative. 

Proof 
-\-mx  -^  {-\-m)  —  -[-X,  because  -\- x  x  (4-^)  =  -\-mx. 
-f-  7nx  -^  (—  7n)  =  —  x,       "        —  X  X  (—  m)  =  +  mx. 

—  7nx  -h  (+  w)  =  —  X,       "        —  X  X  {-\-m)  =  —  7nx, 

—  7nx  -^  (—  w)  =  -f-  ^-       *^        -\-x  X  {—  7n)  — •  —  mx. 
The  condition  to  be  fulfilled  in  all  four  of  these  cases  is 

that   the   product,  quotieiit  X  divisor,  shall  have  the  same 
algebraic  sign  as  the  dividend. 


02  THEORY  OF  ALGEBUAIC  SIGNS. 

EXERCISES. 

Express  the  following  diyisioiis,  reducing  fractions  to 
their  lowest  terms: 

1.  a'—a''b-\-a¥—  ab^-^  +  ab.    2.  «'  —  «V  ^  ax"  -, ax. 

3.   —  ab  —  be  —  ca-\-  abc.  4.  d'b  —  Vc  —  &a  -. rr^V^ 

105.  Rule  of  Signs  in  Fraetions.  Since  a  fraction  is  an 
indicated  quotient,  the  rule  of  signs  corresponds  to  that  for 
division.  The  following  theorems  follow  from  the  laws  of 
multiplication  and  division: 

1.  If  tJie  terms  are  of  the  same  sign^  the  fraction 
is  positive;  if  of  opposite  signs.,  it  is  negative. 

2.  Changing  the  sign  of  either  term  changes  the 
sign  of  the  fraction. 

3.  Changing  the  signs  of  hotli  terms  leaves  the 
fraction  with  its  original  sign. 

4.  The  sign  of  the  fraction  maij  he  changed  by 
changing  the  sign  written  before  it. 

m, 
—  n 


TT.v  A 

.MP 

m       —  m           — 1)1 

T  T?      1                   —                    

n        —  n              n 

Ex. 

2. 

m            —  7)1       — 1)1         m, 

n            —n          n         —  ri 

Ex. 

3. 

m  —  n      11  —  m           in  —  n 

a  —  b         b  —  a             b  —  a 

EXERCISES. 

n  —  in 


Express  each  of  the  following  fractions  in  three  other 
ways  with  respect  to  signs: 

m  —  n  Q         m  —  n  c 

J.    .  /C. .  o. 


c                                        c  )n  —  n 

4.   ^~y .                5.  ^±-^.  6.  i'^i-tr. 

'  m  —  ri                      '  p—  (f  '  p-\-  ^1  —  '' 
Write  the  folloAving  fractions  so  tliat  the  symbols  x  and  y 
,hall  be  positive  in  both  terms: 

x-a                     g    a~x  Q    a-x- h 

'   c  —  y                       '  b  —  y  '  m  ^y  —  c 

10.  _  l:z±,         11.  "^.  12.  -  i^£r  i 

o—y                   b-^  hy  m  +  y-c 


RULES  OF  SIGNS  IJV  DIVISION.  ya 

Aggregate  the  following  fractional  expressions: 

X         2x  3x        X  —  2a a 

ni       —  m       —  7H        —  m        rn ' 

c-  y       c+y       c-2y       2c  -  y 
^**      h  -  /i    "^    -  h  h     ' 

m  —  71        —  m         —  m  —  2a   .    2m  —  a 

lo. :: + 


a  —  n       a  —  71  n  —  a  n  —  a 

106.  General  Remark.  It  is  necessary  to  the  inter- 
pretation of  an  algebraic  result  that  the  positive  sense  or 
direction  be  defined  in  the  case  of  each  symbol  which  admits 
of  either  sign. 

The  understanding  is  that  the  sense  in  which  the  symbol 
is  defined  is  positive. 

Example.  If  we  say,  ^^Let  t  be  the  number  of  days  he- 
fore,^^  we  understand  that  days  before  are  positive  and  days 
after  negative. 

But  if  we  say,  ^^  Let  t  be  the  number  of  days  after y^^  the 
reverse  is  understood. 

107.  Exercises  in  changing  Algebraic  Expressions  into 
Nmnbers.  Compute  the  values  of  the  following  expressions 
when 

a  =  2,  p  =  —  S, 

c  =  5,  r  =  —  S. 

1.  a  -j-p.  2.  a  —p.  3.  b -^  q. 

4:.  b  —  q.  6.  a-\-r.  e,  a  —  r. 

7.  2r  —  3b.  8.  qr  —  cp.  9.  (pq-ac)''. 

10.   {cr-abp)^.  11.   (A]  12.   (- -Y. 

13.   ^'  +  ^~\  14.  c"r-p'q  ^^   pqr-abc 

a"  —  q-P'  '      q-\-r    '  '  pqr-\-abc 

16.   -^.  17    ^^P  ~  ^^^  18    P'-^'"' 


q  —  p  aqp  —  bcr  p  ^  c 

19.  Compute  the  several  valncs  of  the  expression 
ax"^  -\-qx-\-r 
for  iC=:-4,  -3,  -2,  -1,  0,  1,  2,  3,  4. 


CHAPTER   II. 

OPERATIONS  WITH  COMPOUND  EXPRESSIONS. 


Sectiot^  I.    Preliminary  Defiis^itions  and     . 
Principles. 

108.  Aggregate.  A  polynomial  enclosed  between 
parentheses  in  order  to  be  operated  upon  as  a  single 
symbol  is  called  an  aggregate. 

Entire.  An  entire  quantity  is  one  wliicli  is  ex- 
pressed without  any  denominator  or  divisor,  as  2,  3, 
4,  etc. ;  a,  ^,  x^  etc. ;  2aZ>,  2m^,  ah  {x  —  y\  etc. 

Formula.  A  formula  is  an  algebraic  expression 
used  to  show  how  a  quantity  is  to  be  calculated. 

Reciprocal:  The  reciprocal  of  a  number  is  unity 
divided  by  that  number.     In  the  language  of  algebra, 

Reciprocal  of  N  =  -^. 

Function.  An  algebraic  expression  containing  any 
symbol  is  called  a  function  of  the  quantity  repre- 
sented by  that  symbol. 

Example  1.     The  expression  dx'^  is  a  function  of  x. 

Ex.  2.     The  expi-ession  — ^^—  is  a  function  of  x.    It  is 
^  a  —  X 

also  a  function  of  a. 

Degree.  The  degree  of  a  term  in  one  or  more 
symbols  is  the  number  of  times  it  contains  such  sym- 
bols as  factors. 

Examples.  The  expression  abx^y"^  is  of  the  fourth  degree 
in  a^  l  and  x,  because  it  contains  these  symbols  four  times  as 
factors. 

It  is  also  of  the  third  degree  in  x,  of  the  fifth  degree  in 
X  and  ijy  and  of  the  seventh  degree  in  a,  b,  x  and  y. 


PIUJS'GIPLKS  OF  ALGEDllAIG  LANGUAGE.  95 

The  expression  al/x^  is  of  the  fifth  degree  in  b  and  :r,  be- 
cause it  contains  h  twice  and  x  three  times  as  a  factor. 

When  an  expression  consists  of  several  terms,  its 
degree  is  tliat  of  its  liighest  term. 

Principles  of  Algebraic  Language. 

109.  First  Principle.  We  may  tnvploy  a  single 
symbol  to  represent  any  algebraic  expression  what- 
ever. 

The  fact  that  a  symbol  is  meant  to  represent  an  expression 
is  indicated  by  the  sign  =. 

Example.     The  statement 

p  ^  ax  -\-  hy 
means,  we  wa'ite  the  symbol  p  to  represent  the  expression 
ax,  +  lij. 

Second  Principle.  Any  algebraic  expression 
may  be  operated  with  as  if  it  were  a  single  symbol. 

An  expression  thus  operated  on  is  enclosed  in  parentheses 
wlien  necessary  to  avoid  ambiguity. 

As  a  consequence  of  this  principle,  an  algebraic  expres- 
sion between  parentheses  may  be  enclosed  between  other  j^a- 
i-entlieses,  and  these  between  others  to  any  extent.  Each  order 
of  parentheses  must  then  be  made  thicker  or  different  in  form 
to  distinguish  them. 

Third  Principle.  An  identical  equation  will  re- 
main true  wlien  any  expression  or  number  is  written 
in  place  of  each  symbol. 

EXERCISES. 

Let  us  put  P  ^  a  -\-  h', 

Q  =a  -  h; 
and  R  =  ah. 

It  is  then  required  to  substitute  in  the  following  expressions 
a  and  b  in  place  of  P,  Q  and  JL 

1.  PQ.  Ans.   {a-^b){a-b). 

2.  rQR\  Ans.   {a -\- b){a  -  b){ab)\ 

3.  P(P-Q).  Ans.   {a-^b){a-^b-  {a-b)\. 


96       OPERATIONS   WITH  COMPOUND  EXPRESSIONS. 

4.    P{R  -  P),  5.   P(7^  _  0. 

6.  Q{P  -  QR),  Ans.   {a  -  h)  [a -^  h  -  {a  -  h)ah}. 

7.  Q(Q-FE).  8.  P{QR~P), 

110.   Exercises   in   Compound   Expressions.      Compute 
:lie  value  of  the  following  expressions  when 

rt  =  6,  m  =  —  4, 

i  —  ^,  n  =  —  5. 

1.  {(f«  -  b)m  4-  3w{^.         Ans.    (-  3/m  +  3?i)9  =  —  27, 

2.  {vm(^- ^0  4- ^H^  +  ^0 1(^-^0- 

3.  u\b{m  -f  ?^)  —  rt(m  —  n)]. 

4.  Jrr  +  a{h  -  7n)Y  —  \m  +  ni^m  -  u)Y. 

5.  \{a-hY-\-{in-nY\{in-^n). 

0.  (^/,  —  ^  4~  ^^'  ~  ^0^  (^'^  +  ^0- 

7.    {(a  -  m){b  -  ny  -  {a  -  n){lj  -  mY}\ 


Sectio:n^  II.  Cleaeit^g  of  Compound  Pakentheses. 

111.  When  expressions  in  parentheses  are  enclosed  be- 
tween others,  they  may  be  removed  by  applying  the  rules  of 
§§  32  and  33  to  one  pair  at  a  time. 

We  may  either  begin  with  the  outer  ones  and  go  inward, 
or  begin  with  the  inner  ones  and  go  outward. 

It  IS  common  to  begin  with  the  inner  ones. 

EXERCISES. 

1.  Clear  of  parentheses  P  —  {a—  \in  —  (x  —  y)] }. 
Solution,     Beginning  with  the  inner  parentheses  and  ap- 

l)lying  the  rule  of  §  33  to  each  i3air  of  parentheses  in  succes- 
sion, the  expression  takes  the  following  forms: 
P-\a-  {m-x-{-y)] 
—  P  —  \a  —  nfi -\- X  —  y] 
=  P—  a  -\-  m  —  X  -\-  y.       Ans. 

2.  x—{'ila  —  x—{a-hx)-{-Za-^j)\. 
Solution,     Eemoving  the  inner  parentheses,  we  have 

x  —\^a  —  X  —  a  -\-bx-\-^a  —  x\ 
=^  X  —  '^a  -{-  X  -\-  a  —  ^x  —  'da  -\-  X 
:=  —  2x  --  4ca.    Ans. 


GENERAL  LAWS  OF  MULTIPLICATION.  97 

3.  a  +  {  -  [a  -  2b)  +  (2a  -  Z/)  -  (-  ^a  -  3b)}. 

4.  -(x-  a)  -  {2x  -  3a)  +  {6x  +  ^a). 

5.  |w-  [^y?  -  (m-2m)]}. 

G.  2?/2.2:  -  J3m2;  -\-p!/  -  (omx  -  2py)  -  {- ^py  -r  t^x)]. 
7.    _  {{]]jy  _j-  '^mu)  +i  -  2%  -  {mu  -  3/;?/)  +  5/^^  -f  2?/iw  j. 

9.  a  -  X  -{a-x-{a-\-  x)  -  (x  -  a)  J . 

10.  3ax  -  2hy  -  {ax  -  by  -  [ax  -by  -  {by  +  ax)]}. 

11.  X  —  {2m  -\-  (5a-  —  a  —  b)  —  (7rt  +  2b) }. 

12.  -  (367/  +  2mz)  -  {2r/y  -  {3mz  -  2cy)]. 

13.  p-q  - {2p  -  3q  -  {^p  +  ^(j)  +  (G^^  -  ^) }• 

14.  p-^qJr{-3p  -  4r/  -  {2p  -  Try)  +  (3;^+  4^)}.       • 


Seotiox  hi.    Multiplk  ation. 
General  Laws  of  Multiplication. 

112.  Law  of  Commutation.  Multiplier  and 
wnltiplicand  may  he  interchanged  witliout  altering 
the  product. 

This  law  is  proved  for  whole  numbers  in  the  folloAving 
way:  Form  several  rows  of  quantities,  each  represented  by 
the  letter  a,  with  an  equal  number  in  eacli  row,  tlius: 


a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 

Let  m  be  the  number  of  rows,  and  n  the  number  in  eacli 
row.     Then  counting  by  rows  there  Avill  be 

n  X  tn  quantities. 
Counting  by  columns  there  will  be 

m  X  n  quantities. 
Therefore  m  X  n  =  7i  X  m, 

or  nyn  =  7mi. 

Remark.     This  is  called  the  kw  of  eoinmntation,  from  commuto,  I 
cxchaiiffe. 


98       0PEnAT102s'S  WITH  COM  POUND  EXl'RESSlOyS. 

113.  Law  of  Association.  When  there  are 
three  factors,  m,  n  and  a, 

7n{na)  =  {vin)a. 
Example.      3  x  (5  X  8)  :==    3  x  40  =  120; 
(3  X  5)  X  8  =  15  X    8  =  120. 

Proof  for  Whole  Numbers.  If  a  in  the  above  scheme 
rei)resents  ii  number,  the  sum  of  each  row  will  be  na.  Be- 
cause there  are  m  rows,  the  whole  sum  will  be  m(na). 

But  the  whole  number  of  as  is  mn.     Therefore 

7ti{na)  =  {mn)a. 

Remark.  This  is  called  the  law  of  association,  because  the  middle 
factor,  n,  is  associated  first  witli  a  and  then  with  m.  • 

114.  The  Distributive  Law.  The  product  of 
a  polynomial  hy  a  factor  is  equal  to  the  sum  of 
the  products  of  each  of  the  terms  hy  the  same  factor. 
That  is, 

m{p  -[-  q  -\-  r  -X-  etc. )  =  mp  -\-  mq  -\-  mr -{-  etc.         (1) 

Proof  for  Wliole  Numhers.  Let  us  write  each  of  the 
quantities  j9,  q,  r,  etc.,  m  times  in  a  horizontal  line,  thus: 

p  -{■  p  -\-  p  -\-  etc. ,  m  times  =  mp. 
q  -\-  q  -{-  q  -\-  etc.,  m  times  =  mq. 
r  -\-  r  -\-  r  -\-  etc.,  m  times  =  mr. 
etc.  etc.  etc. 

If  we  add  up  each  Ycrtical  column  on  the  left-hand  side, 
the  sum  of  each  will  ha  p  -{-  q -\-  r  -\-  etc.,  the  columns  being 
all  alike. 

Therefore  the  sum  of  the  7n  columns,  or  of  all  the  quanti- 
ties, will  be 

m{p  -\-  q  -\-  r  -^  etc.)t 

The  first  horizontal  line  of  7^'s  being  7np,  the  second  mq, 
etc.,  the  sum  of  the  right-hand  column  will  be 
Trip  +  mq  -\-  mr  +  etc. 

Since  these  two  expressions  are  the  sums  of  the  same 
quantities,  they  are  equal,  as  asserted  in  equation  (1). 

Remark.  This  is  called  the  distnbutive  law,  because  the  multiplier 
m  is  distributed  to  the  several  terms  jo,  q,  r,  etc.,  of  the  polynomial. 


FOiiMATioy  OF  riionucTS.  99 


Forinatioii  of  Products. 

115.  Products  of  Aggregates.  Expressions  between 
parentheses  may  be  multiplied  and  the  parentheses  removed 
by  successive  application  of  the  principles  of  §  38,  and  of  the 
distributive  law. 

EXERCISES. 

1.  a\m  —  n{p  —  q)]. 

Solutmi.     Apj)lying  the  rule  of  §  38,  we  have 
a{m  —  np  -\-  nq]  =  am  —  anp  -\-  anq.     Ans. 

2.  m{ax  —  h(y  —  z)]. 

3.  p{qm-  l)\a-h)]. 

4.  PxP  +  qip-  q)\- 

5.  m^\ax  —  '6{iix  —  hy)  -\-  5{a.'^x  —  Ify)]. 

6.  n{p-\-q[x-Yy(a-b)\\. 

7.  r{t-m[x-y{p-q)Y^.       ' 

8.  r{l-\-r[l^r(l  +  r)]]. 

9.  rjl- r[l  -  r(l  -  r)]}. 

10.  a-[-x\h^  x[c  +  x{cl  -f  x)'\  ], 

11.  d-x{c-  q:\1)  -  x{a  -  .r)] }. 

12.  a^\m{b  —  a)'}i  —  n{a  —  b)m}ax^. 

13.  pq{a{b  -  c)  +  b{c  -  a)  +  c{a  -  b)]pr. 

14.  pq\a{b  -  c)-\-  2b{c  -  a)  +  3r(a  -  b)\pr. 


15.^]-^^- 
n  [       m 


w  I         n 
n  \        'ill 


116.  Arrangements  according  to  Powers  of  a  SyinboL 
We  may  collect  all  the  coefficients  of  each  power  of  some  one 
symbol,  and  affix  the  power  to  their  aggregate,  by  the  process 
of  §  26. 

Def.  When  a  polynomial  is  arranged  according  to 
powers  of  a  symbol  it  is  called  an  entire  function  of 
that  symbol. 

Example.     Arrange  according  to  powers  of  x, 
x{iax  -  2b)  -  b(3x'-  F)  +  (a  -  4b)x  -  nax'-\-  ic. 

Solution.     Clearing  of  i)arentheses, 

4:ax'  -  2bx  -  3^.^-'  +  ^'  +  ax  -  ibx  -  dax''  -\-  4.c. 


100     OPERATIONS  WITH  COMPOUND  EXPRESSIONS. 

Taking  the  coefficients  of  the  several  powers  of  x,  we  find 
them  to  be: 

Coefficients  of  x^  =  4:a  —  3b  —  3a  =  a  —  3b; 
Coefficients  ot  x   =  —  2b  -{-  a  —  4:b  =  a  —  Qb; 
Term  without  2;    =  b^  -\-  4c. 
Therefore  the  expression  is  equal  to 

(a  -  3b)x'  +  («  -  (jb)x  +  Z*'  +  4c. 

EXERCISES. 

Arrange  the  following  expressions  according  to  the  powers 
of  x  or  other  leading  symbol: 

1.  (a  +  x)x''  —  {b  —  x)x^  -\-  ex  —  d. 

2.  {a  -\-  bx)x  +  (c  +  dx)x'^  —  a. 

3.  i^nx^  —  n  —  2))x  -\-  [2])  —  q)x  —  ax^  -j-  ^^^  —  ^^' 

117.  By  combining  the  operations  of  the  two  j)receding 
articles  entire  functions  of  one  symbol  may  be  expressed  as 
entire  functions  of  another  symbol. 

EXERCISES. 

Arrange  the  following  polynomials  according  to  poAvers 
of  x\ 

1.  {2x  -  3x')f  -{x^  2x')y'  +  {3x'  -  2x'  -  a)y. 

2.  (ax^  +  bx^)y  -\-  y^'lax'  —  (3a  +  y)x}  —2xy\ 

3.  {a  +  a^xy  +  x^)a'^  -\- {b  -{-  ex -{-  mx^)a  -\-  2cx  —  nx^. 

4.  aimix'  -y)-  n  {x'  -y')\-{-  b{y'  +  y'x  +  yx'  +  x'). 

5.  (a'  +  2b'x  +  2cx'  +  x')xy  -  7n{b'  +  2c'x  -  Mx"), 

Multiplication  of  Polynomials  by 
Polynomials. 

118.  Let  us  consider  the  product 

{a^b){p^q  +  r). 
liegarding  n^  +  ^  'is  ^  single  symbol,  the  product  by  §114  is 
(a  +  b)p  +  (^  +  b)q  +  (a  +  b)r. 
But  {a  -f-  b)p  z=  ap  -\-  bp\ 

(a  +  b)q  =  aq  -j-  bq; 
(a  +  b)r  =  ar  -\-  br. 
Therefore  the  product  is 

^1^  +  ^P  +  ^(/  ~\-  M  ~V  ^^^  +  ^^« 


MULTIPLICATION  OF  POLYNOMIALS.  101 

It  would  have  been  shorter  to  first  clear  the  parentheses 
from  {a  -{-h),  putting  the  product  into  the  form 
<P  -\-  q-{-r)  ^h{p  ^-  q  +  r). 

Clearing  the  parentheses  again,  we  should  get  the  same 
result  as  before. 

We  have  therefore  the  following  rule  for  multiplying 
one  polynomial  by  another. 

119,  KuLE.  Multiply  each  term  of  the  multiplicand  by 
each  term  of  the  multiplier,  and  add  the  products  with  their 
proper  algebraic  signs. 

EXERCISES. 

1.  (m  —  n)  {p  —  q). 
Solution,      (m  —  n)p  =       mp  —  np; 

(m  —  n)  X  (—  q)  =  —  mq  -[-  nq. 
.*.  product  =  7np  —  nj)  —  mq  +  7iq, 


2. 
3. 

4. 

5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 


a  -  h)  {x  -  y  ^  z). 

a  J^  I)  (x  -\-  y  -  z). 

a  +  l)  {x-\-  y)  -\-  {a  -i)  (x  -  y). 

m  —  n)  {x  -\-  y)  —  {m  +  u)  {x  —  y). 

P  +  q)  (^^  -  %)  +  {P  -q)  {ax  +  hy). 

m  -  q)  {mx  +  qy)  -\-  {m  -  q)  {qx  -  my). 

7nx  +  ny)  {my  —  nx)  +  {mx  —  ny)  {my  +  nx), 

2a  —  h)  {2a  —  35  +  4c). 

X  -l){x-  2). 


X  —  a)  {x  —  I)  —  {x  -\-  a)  {x  -^^  l\ 
2a  -  2)  («  +  1). 
1  +  m)  (2  —  m). 

u.  (^  +  « )  f  1  +  ™  +  f.). 

\n    '   mj  \      ^    n    ^   ml 

"■&+■)£-.)■ 

\?r        nj  \m^       mj 

17.  {a'-^a-\-l)  {a' -a -]-!). 

18.  {a'  +  «^  +  F)  {a'  -ab  +  h"). 


102       operations' S  WITH  COMPOUND  EXPRESSIONS. 

120.  The  beginner  will  often  find  it  convenient  to  write 
the  multiplier  under  the  multiplicand,  as  in  arithmetic.  In 
writing  the  separate  products,  like  terms  may  then  be  written 
under  each  other  in  order  to  be  added  up. 

Example,     {a"  +  <^'  +  d"  —  ab—lc  —  ca)  (a-]-b-{-  c). 
Work: 

(i"  -j-  ^'^  +  ^^  —  ab  —  be  —  ca 
a  -\-b  -\-  c 

a^  -\-  aV  +  ac"  —  a'b  —  abc  —  a^c 

-  ab''  a^b  -  abc  +  ^'^  +  be"  -  b\ 

—  ad"  —  abc  +  aj'c  —  b&  +  Wc  +  ^'^ 

"a^  -Mbc  +^^  +c' 

Hence  product  =  a^  -^  ¥  -\-  &  —  Zabc. 

EXERCISES. 

In  the  following  exercises  arrange  all  the  terms,  both  of 
multiplicand  and  multiplier,  according  to  the  powers  of  the 
leading  symbol. 

Multiply: 

1.  x'^  -{-  ax  -\-  a''  by  x  —  a. 

2.  x^  +  ax'^  -\-  a'^x  -f-  a^  l)y  x  —  a. 

3.  x^  —  ax""  -\-  a^x"  ~  a^x  +  a'  by  x  +  a, 
i.  1    —  X  -{-  x""  —  x^  -f  x^  by  1  -f-  X. 

5.  1    -  2x  -\-  dx'  -  4:x'  by  1  —  x  +  x\ 

6.  x'  -  dx'  -{~3x  -  1  by  x'  -  2x  +  1. 

7.  X*  ^  x'  -\-l  hjx*  -{-  x'  -  1. 

8.  a'  +  2a''  -{- 2a -{- 1  by  «'  -  2a  +  1. 

9.  a'  -  2a''  -{-  2a  -  1  by  a''  +  2a  +  1. 

10.  ^-  _  2  -  +  3  bv  -^  +  2 3. 

a  a  ^  a  ft 

11.  a'  -  2a'  +  3a''  -2^  +  1  by  a'  +  2a''  +  2^  +  1. 

12.  (x  -  3)  {x  -  1)  (x  +  1)  (a;  +  3). 

13.  n{n  -  1)  (n  -  2)  (?i  -  3). 

14.  {x  +  «)  (:r^  +  a')  (x  -  a). 

15.  (a''  +  «  +  1)  (^^^  -  rt  +  1)  (a*  -  a'  +  1). 
10.  a''  -{-  Aax  +  4a;'  Ijy  a^  -  Aax  +  4:x\ 

17.  2./;'  +  4.a;'  +  8.t  -f  16  by  3x  -  0. 

18.  a'  +  <^'  +  6'^  +  «^  +  ^t;  -  ca  hy  a  ~b  ^  c. 


SPECIAL  FORMS  OF  MULTIFLUJATION.  103 


Special  Forms  of  Multiplication. 

131.     Square  of  a  Binomial,     To  find  the  square  of  a 
binomial,  as  a  -\-b.     We  multiply  a  -\-  b\}^  a^  b. 
a(a  -\-  b)    =  a""  -\-    ab 
h{a  -\-b)    ::^  ab  4   b' 

(a  +  b)  {a  +  b)    =  a'~'-^2ab  +  b' 
Hence  (a  +  by  =  a'  +  2ab  +  b\  (1) 

Def.  When  a  monomial  expression  is  reduced  to 
the  algebraic  sum  of  several  terms,  it  is  said  to  be  de- 
veloped. 

EXERCISES. 

Prove,  by  computing  both  members,  that 
(2  -{-  Sy  =  2'  +  2  .  2  .  3  -}-  3\ 
(4  +  3)^  =  4^  H-  2  .  4  .  3  +  3\ 
(2  +  5)^  =  2^  +  2  .  2  .  5  +  b\ 
Write,  on  sight,  the  developed  values  of  ihc  following  ex- 
pressions: 

1.  {m-^7iy.  2.  {m-\-2ny.             3.   (2m  ^  n)'. 

4.  (ax-]-byy.  5.  (ax -^  2byy.           6.   (2r^.c  +  %)^ 

7.  K  +  ir.  8.  (ab-i-iy.               9.   (2ab-^iy. 

10.  (Z»^+^r.  11.  (b'-i-2by.  12.   (2Z»^  +  ^)l 

13.  (m'-\-ny.  14.  (m=^  +  «;^)^  15.   (^m^  +  ^^^^^)^ 

10.  (b-{-2y.  17.  (^  +  3^.  18.   (^  +  4)1 

19. 


.(.+i)-.    ^.(,.+,f^    „,(|.+^g. 


22, 


123.  We  find,  in  the  same  way, 

(a  -  by  =  a'  -  2ab  +  b\  (2) 

This  and  the  preceding  form  may  be  expressed  in  words 
thus: 

Theorem.  The  square  of  a  hinoiuial  is  eqaal  ht  the  sinn 
(f  fhe  sq/farrs  of  ils  f irii  Icnits  plus  or  minus  hricc  llwir  pm- 
(lurL 


104      OPERATIONS  WITH  COMPOUND  EXPRESSIONS. 


EXERCISES. 


Show  that  the  preceding  formula 

gives 

correct  values  of 

(5 

-  3)s  (5  -  ly, 

Develop: 

and  (9  -  4.)\ 

1. 

{m  -  7i)\ 

2.  {m-2n)\ 

3.   {2m-n)\ 

4. 

{a  -  1)\ 

5.  (a-2y. 

6.   {a  —  d)\ 

7. 

{ah  -  x)\ 

8.   (ab  -  2xy. 

9.   {2ab  -  xy. 

10. 

(ax  -  hy)\ 

11.  {ax-2byy. 

12.   {2ax  -  bi/y. 

13. 

hn       nV 
\2  "  5^)  ' 

\a       2b  J 

15.    f  -  !L^. 

\2a       0  J 

123.  Product 

of  the  ^Sum  and  j 

Difference.     Let  us  find 

the  product  of  a  +  ^  ^J  ^^  "  ^• 

a{a  -\-  b)  =  a""  +  ab 
-  b\a  -\-b)  ^       -  ab  -W 

Adding,        (a  ^b){a-b)=^  a'  ~-l)\  (3) 

That  is: 

Theorem.  Tlie  product  of  the  sum  and  difference  of  two 
numbers  is  equal  to  the  difference  of  their  squares. 

Remark.  The  three  preceding  forms  should  be  carefully  memorized 
by  the  student,  owing  to  their  constant  occurrence. 

EXERCISES. 

Form  the  following  products  on  sight: 

1.  {m  —  n)  (m  +  n).  2.  (m  —  2n)  (m--{-  2n). 

3.  (2m -{-u)  (2m  —  n).  4.  (ab  +  c)  (ab  —  6-). 

5.  (ab  +  2c)  (ab  -  2c).  6.  (2ab  +  c)  (2ab  -  c). 

7.  (ax  +  1)  (ax  -  1).  8.  (ax  +  2)  (ax  -  2). 

9.  (ax  +  5)  (ax  —  5).  10.  {ax-\-by)  (ax  —  by). 

11.  (a'x  -  b'y)  (a'x  +  Fy).  12.  («V  -  bY)  [ct'x'  +  b  y'). 


2a      2b    \2a      2b 


15.   {a"  +  \)  [f  -  ^).  16.   (a-^b^c)(a^b-  r). 

Ans.  (16).     (a  +  by  -  &  =  a^ -\-  2ab  ^V  ^  c\ 
17.   (x^y-^z)(x-\-^i-  z).      18.   (1  +  ?^  +  w')  (1  +  ^^  -  n''). 
19.   {ij^a+2b)  (l^a-2b).        20.   (a-^2b-\-2c)  (a-\-2b-2c). 
21.   (a-\-b^ab)(a-\-b-ab).22.    (3 -}- 4  -  5)  (3  +  4  +  5). 


SPECIAL  FORMS  OF  MULTIPLICATION.  105 

134.  Because  the  product  of  two  negative  factors  is 
positive,  it  follows  that  the  square  of  a  negative  quantity  is 
positive. 

Examples.       {—  aY  =  a'  =  (+  af. 

(b  -  ay  =  a'  -  2ab  -\- h'  =  {a  -  h)\ 
Hence 

The  e:qjression  a'  —  2ab  -f  F  is  the  square  both  of  a  —  b 
and  of  b  —  a. 

125.  We  have         —  a  X  a  =  ~  a". 
Hence 

Tlic  product  of  equal  factors  with  opposite  signs  is  a  ncrja- 
tice  square. 

Example.  —  {a  —  b)  {a  —  b)  =^  —  «'  -f  '^(d)  —  h\ 
which  is  the  negative  of  (2).  Because  —  [a—  b)  =  b  —  a, 
this  equation  may  be  written  in  the  form 

(b-a)  {a-b)  =  -a'-i-  2ab  -  b\ 
which  is  readily  obtained  by  direct  multiplication. 

EXERCISES. 

Form  the  products: 

1-  (^'  -y){y-  ^)-  ^-  {^  +  y)  (y  -  ^)- 

3.    -  a'  X  d\  4.    {a  -  2b)  (2b  -a). 


5. 


S"^)(^~3*        ^'  (^^•'^•-%)(%-^^-'-). 


136.  /Squares  of  Trinomials.     Let  us  form  the  square  uf  . 
a  -\-  b  -[-  c. 

a{a  -\-  b  -}-  c)    =  a""  -{-  ab  -\-    ac 

b(a  +  b  -^  c)    =  b'  -{-  ab  +    be 

c(a  -\-  b  -{-  c)    =  c^  +    ac  +    be 

Sum  =~(a  +  b-\-cy  =  a'-j-b'  +'7'+la^  +  2ac  +  2bc. 

Hence 

Theorem.  The  square  of  a  trinomial  is  equal  to  the  sum 
of  the  squares  of  its  terms  plus  twice  the  product  of  each  pair 
of  terms  tahm  tico  and  two. 

This  theorem  applies  to  a  polynomial  of  any  number  of 
terms, 


lOG      OPE  RATIO  yS  WITH  COMPOUND  EXPRESSIONS, 

EXERCISES 

Develop  the  expressions: 

1.  (^a-b-  c)\     Ans.  a'  +  b'  +  c'  -  2ab  -  2ac  +  2bc. 

2.  (a  -^b  -  cy.  3.   {x  -  2a  +  b)'. 
4.   (a--ab-{-by.  5.   (a'-2ab  +  by. 
6.   (;m^  _!_  ,,^  _  3^^)^  7.   (1  +  e+eT- 

8.   (.c^  _  :i,- +  l)^  9.   {2x'' -dx  +  4.y. 

10.   (r^/^  +  ^^  +  «/)'.  11.   {(^"b  +  Z^V'  +  c'ay,      ' 


14.    («_l  +  ^y.  15.    (iJ^.c  +  x'-^xy 


Section  IY.    Division  op^  Compound  Expressions. 

Division  of  one  Polynomial  by  Another. 

127.  Case  I.  When  there  is  only  one  algebraic  symbol 
in  both  dividend  and  divisor. 

Rule.  1.  Arrange  both  dividend  and  divisor  according  to 
the  2)oivers  of  the  symbol. 

2.  Divide  the  first  term  of  the  dividend  by  the  first  term 
of  the  divisor,  and  write  down  the  quotient. 

3.  Multiply  the  whole  divisor  by  this  quotient,  and  sub- 
tract the  pi'oduct  from  the  dividend.  • 

4.  Divide  the  first  term  of  the  remainder  by  the  first  term 
(f  the  divisor,  and  repeat  the  process  until  the  divisor  ivill 
no  longer  divide  the  remainder. 

Rematik.  Unless  we  introduce  fractions,  the  dividing  process  neces- 
sarily stops  when  the  degree  of  the  remainder  is  less  than  that  of  the 
divisor. 

Example.     Let  us  perform  the  division 

dx'  -  4:x'  4-  2x'  -\-'Sx  -  1  -^  {x'  -  x-\-  1). 

We  first  find  the  quotient  of  the  highest  term  of  the  divisor  x^  into 
the  highest  term  of  the  dividend  3.v*,  wliich  is  3^^  ^^^.  ti^^.^  multi- 
ply the  whole  divisor  by  the  quotient  S.r^.  and  subtract  the  product  from 
the  dividend.  We  repeat  the  process  on  the  remainder,  and  continue 
doing  so  until  the  remainder  has  no  power  of  x  so  high  as  the  highest 


DlVlt^WN  OF  ONK  POIANOMIAL  liT  ANOTJlEll    lo7 

term  of  the  divisor.     The  work  is  most  conveniently  arranged  as  fol- 
lows : 

Dividend.  Divisor. 

^x'-  ^x'-\-  2x''-\-^x  -  1  !  :7--a;  +  l 

3.r2  X  divisor,         'dx* —  dx^-\-  ^X^  ^6x' —  X  —  2      Quotient. 

First  remainder,  —     X^ —     X^-\Sx  —  i 

-.rX  divisor,  —     X^-{-     X^  —    X 

Second  remainder,  —  2x^-\-4:X  —  1 

-2X  divisor,  —  '2x^-\-2x  —  2 

Third  and  last  remainder,  llx  -\-  1 

The  division  can  be  carried  no  farther  without  fractions,  because  x^ 
will  not  go  into  x.  We  now  apply  the  same  rule  as  in  arithmetic,  by 
adding  to  the  quotient  a  fraction  of  which  the  numerator  is  the  remain- 
der and  the  denominator  the  divisor.     The  result  is 

x'—  X  -\-  I  X-—  x-\-l         ^  ^ 

This  result  may  now  be  proved  by  multiplying  the  quotient  by  the 
divisor  and  adding  the  remainder. 


EXERCISES. 

Execute  the  following  divisions,  and  reduce  the  quotients 
to  the  foi'm  {a)  when  there  is  any  remainder: 

1.  x'—  2a;'+  %x-l-^x-l. 

2.  3a;'+  Qx'-  2x'-  2x  ~^  x'  -  2. 

3.  48a:'-  IGx'-  32x  +  50  -:-  2a:  ■-  3. 

4.  a'-\-l-^a-{- 1. 

5.  a^-\-  a^-\-  a  -\-l  -^  a-{  L 

6.  a'-^  a'-{-  2a-{-'i  -^a'  -[- 1. 

8.  (y«+  122/  +16)0/  H-  12.^/  -  16)  -  ^  -  3. 

9.  {f-2y-^l){if-dy^2)-^{y-l){f-^   2;/4-i). 

10.  y*-  2y'-\-  St/+4:y  ^  y'-  y  -|-  1. 

11.  (,/+8)0/-4)0/+l)-2/'-%+l> 

12.  (tf^l)(y^J^^)(y-4.)---f+y-^2. 

13.  (rt'^+3)(^-,1)K-5)-f-«-l. 

14.  (^^_2)(V-4)(r?,-8)  -^r?'^+2«-fl. 

15.  a{a'-\~  1)K+  a){a  -1) -^  a''-2a\  ^^ 


108     OPmATtONS  WITH  COMPOUND  KXPUESSlONf^. 

128.  Use  of  Detached  Coefficients.  In  dividing  by  the 
preceding  method,  there  is  no  need  of  repeating  the  symbol 
after  each  coefficient. 

Example,     x'—  4:x'-\-  Zdx  —  11  ~  2^'+  2^-1. 

X*         X^  X^         X^  X^         X 

-fl_4  0  +  35-111  +  1  +  2-1 


+  1+2-1  +  1^'  -  62;  +  13 


.     -  6  +     1  +  35 

-  6  -  12  +    6 

+  13  +  39  - 
+  13  +  26  - 

11 
13 

+    3+    2 
Here  the  power  of  x  which  is  written  above  each  column  of  coeffi- 
cients is  to  be  understood  as  if  repeated  after  each  coefficient  below  it . 
Result:  Quotient,      x^  —  Qx  -{-  13. 
Remainder,  Zx  +  2. 

EXERCISES. 

1.  a^-  2a'+  4«  -  8  -^  ft  -  2. 

2.  2a*+  6«'+  12a  -  3  -^  «'-  2ft  +  2. 

3.  2ft*-  2ft'+  6ft  -  6  -^  a'-  1. 

4.  3a:'+  Qx'-  lix'-  x'-  11a:  +  15  -^  x'+  2x  ~  5. 

5.  x'-  2.t'+  4:x'^2x  -b~-  x'-  2x  +  5. 

129.  Case  II.  When  there  are  several  algebraic  sym- 
bols in  the  dividend. 

Rule.  Arrange  the  terms  of  the  dividend  and  divisor  ac- 
cording to  the  poicers  of  some  one  symhol,  preferring  that  sym- 
bol which  has  the  greatest  number  of  poivers. 

Then  proceed  as  in  Case  /. 

Example  1.  Divide  a:'  +  Zax^  +  '6a^x  +  ft'  by  a:  +  a. 

OPERATION. 

x^  Ar  Sfto;'  +  3ft'^a;  +  ft"  \x  -\-  a 

x'  +    ax^  x'  +  2ax  +  a' 


2fta;'  +  3ft'a: 
^ax"  +  ^a'x 


ft^r  +  ft® 
a^x  +  ft' 
~V      0 


DIVISION  OB'  ONE  POLYNOMIAL  BY  ANOTHER.    109 

Ex.  2.  Divide  x""  —  ax^  -f  a{l)  -\-  c)x  —  ahc  —  hx"—  cx^—  hex 
hj  X  —  a. 

Arranging  according  to  §  116,  we  liave  the  dividend  as  follows: 

x^  —  («  +  J  +  c)x^  -\-  {ab  -\-  be  -{-  ea)x  —  abc  \x  —  a 

x^  —  ax"^  x'—  {b-\-c)x  -j-  be 


(b  -f  c)x''  -f  (ab  -\-  be  -\-  ea)x 
(b  -\-  c)x^  +  (^^  +  ^^)^ 

bex  —  abc 
bcx  —  abc 


0  0 


4. 

x'  +  o\ 

7. 

x'  -  c\ 

10. 

x"  -  e\ 

EXERCISES. 

Divide  the  following  binomials  hj  x-{-  c  and  hj  x  —  c\ 

1.  x""  +  e\  2.  x'  +  e\  3.  x'  +  e\ 

5.  a;"  +  e\  6.  a:^  +  e\ 

8.  ^'  -  c^  9.  x'  -  c\ 

11.  x'  -  c«.  12.  :z:^  -  c\ 

13.  Divide  a;'  +  (^  +  c)^:'  +  ^>c^  +  aix""  -\-  bx  -^  ex -{-  be) 
by  .T  +  a,  then  by  a;  +  Z>,  then  hy  x  -{-  c. 

14.  Divide    a'  +  J*  +  c*  -  2{a'b'  +  ^V'  +  cV) 
by         (a-^b-{-  e)(a  -{- b  —  e){a  —  5  +  c){a  —  b  —  c). 

Divide  by  the  four  factors  of  the  divisor  in  succession. 

15.  Divide  {ax  +  bf  -[-{a-  bx)'  by  x'  +  1. 

16.  Divide  x*  +  Ua*  by  x""  —  4«.r  +  8r?'. 

17.  Divide  4.a'  +  Sa'^Z*  -^  b'  -  da'  -\-3ab  -  Ihja  +  b -I. 

18.  Divide  x'-\-  y'-\.  2x\f  by  (a;  +  iff. 

19.  Divide  2:"+  if—  2.<?y''  by  (.'T  -  y)\ 

20.  Divide  (a:-2?/)(.T^-32:^.?/+5a;/-3.y'')by:?^»-3:?;y+2?/». 

21.  Divide  ««+  «*Z>*+  Z>«  by  (a"-  ab  +  Z'')(«''+  «5  +  V), 

22.  Divide  («  +  2Z>)  «'-  (Z>  +  2«)^'  by  r?  -  Z». 

23.  Divide  {x^-\-  a')(a:'+  w^  +  a')  bya;*+  aV+  «♦. 


110     OPrntATIO^S  WITH  COMPOLWJJ   i-:x rUK.s.SlU^^S. 

Division  into  Prime  Factors. 

In  the  following  exercises  we  apply  the  definitions  and 
processes  of  §§  50  and  51  to  compound  expressions. 

130.  Difference  of  Tivo  Squares.  By  §  123  the  difference 
of  two  squares  is  equal  to  the  product  of  the  sum  and  differ- 
ence of  their  square  roots,  which  sum  and  difference  are  there- 
fore the  factors. 

EXERCISES. 

Factor: 

1.  a^-h\  Ans.   {a  J^h)[a -h). 

2.  7}f-  n\  3.  ^nf-  n\ 

■      4.  ^m'-  9^^  5.   IG/-  25f/'\ 

6.  x^-l.  7.  x'-l, 

8.  a;'— rt'.  Ans.  {x^-^  a''){x''—  a"). 

9.  x'-  1.     • 

10.  ^x'-  4. 

11.  x'-a\      Ans.  {x''-^ce){x'-a'')  =  {x''-^a:'){x-^a){x-({). 

12.  x'-  1. 

13.  Ua'-V. 

14.  m^x  —  n'x.     Ans.   {iif—  n^)x  =  (w  +  n)i(m  —  n)x. 

15.  {a'-  lf)y. 

IG.  {nf-  7i'){x-]-y). 

17.  {m''-n'')\x''-if). 

18.  x^—xif.  Ans.  x{x  -\-  y){x  —  y)- 

19.  a'-4:ax\  20.  9m'- 4mV. 

1  14 

21.  a'-~  22.   -;,-  -  jj-. 

a^  a        h 

131.  Perfect  Squares.  By  §§  121  and  122  a  trinomial 
consisting  of  the  sum  of  two  squares  j9??^5  or  minus  twice  the 
product  of  their  square  roots  is  the  square  of  a  binomial,  and 
may  therefore  be  divided  into  two  equal  factors. 

EXERCISES. 

Factor: 

1.  x:'-2x-\-  1.  Ans.   {x  -  ly  or  (1  -  x)\ 

2.  x"  -4:X-^  4.  3.  a"  +  4.ax  +  4.x\ 
4.  a"  +  ^a  4-  9.  5.  w'  -  2w'.'r  +  x\ 
6.  2:^  +  lOo;^  +  25.                       7.  49F+14A:  +  1. 


Divisions  L\T(j  VlUMh:  FAuTOltS.  m 

8.   x^  +  'Hax'  +  era:     Ans.  a'{;A'  +  a)\ 
0.  a'  —  4:a'w  +  ^am\ 

10.  2;^  +  2«a;  +  a'  -  b\ 

Ans.     {x  J^ay-F  =  (x-}-a  -f-  Z»)(rr  +  ^  -  b). 

Note.    Here  we  combine  the  methods  of  this  and  the  pieceding  sec- 
tious, 

11.  x"-  -2ax-{-a'-h\ 

12.  m'  —  4w  -  x'  +  4. 

13.  3a'  -  6«m  +  3wi'. 

Ans.  ^ce-2am-^m'')=^a-m)\ 

14.  Sft'^  -  6aZ>  +  U\ 

15.  3w' -  6«&  +  3^^' -  4^1 
IG.  5m'+5?i'^-20//-10ww. 

17.  2A'  +  8F-  8//1-  -  18AU-\ 

Ans.  2(7/.  +  2h  +  37/y[;)(7i  +  21c  -  37/y^). 

18.  3/-12p^+12^'^-27r?'y. 

19.  ax'-\-2(i'x-\-a'  -^a\ 

20.  wijt?'  +  4m>  +  4^//'  -  1G?»>*. 

Ans.   ^6'  +  ly. 


21. 

'^  +  ^  +  \. 

22. 

c^-^  +  l- 

24. 

"'\       4  +  4't 

20. 

a'       2ac        & 
111^      mil       n^' 

23. 

25. 

27. 

132.    Oilier  Trinomials.     If  we  perform  the  multiplica- 
tion 

(^  +  ci){x  +  ^'), 

we  find  as  the  result 

x"  +  (a  +  2')^  -y  cib  =  {x  +  rO(:?^'  +  ^). 
We  see  that  in  the  trinomial 

(1)  The  coefficient  of  x  is  the  sum  of  a  and  ^,  and 

(2)  The  term  independent  of  x  is  their  p^^oduct. 
Hence  if  in  a  trinomial  of  the  form 

0^  -\-  px  -\-  q 
we  can  find  two  numbers  whose  sum  is  p  and  whose  product  is 
q,  the  trinomial  may  he  factored. 


112    0PML4.TION8  WITH  COMPOUND  EXPRESSIOI^S. 

Example.  x^  +  bx  +  6. 

Here  we  find  by  trial  that  2  and  3  are  sucli  numbers  that 
2  +  3  =  5  and  2  X  3  ==  6.     Hence 

x"  -]-bx  -\-  Q  =  [x-]-  2)(^  +  3) 

EXERCISES. 

Factor: 

1.  x""  -\-^x-ir  2.  2.  x''-\-^x^  3. 

3.  x'-\-1x-\-  10.  4.  c'^  +  6c  +  8. 

5.  x'  +  W  +  12.  6.  c'  +  Sc'  +  12. 

7.  m"  +  9m'  +  14.  8.  m'  +  8m'  +  15. 

9.  ax"  +  "iax  +  12«.         Ans.  a{x  +  4)(a;  +  3). 

10.  rri'x''  -\-  12mx  +  35. 

11.  !!^  +  i3!!^  +  40.  12.   ^;  + 11^  +  24. 

71  71  C  C 

133.  If  the  second  term  of  the  trinomial  to  be  factored 
is  negative  and  the  third  positive,  both  the  quantities  a  and 
h  must  be  negative. 

For  the  product  being  positive,  the  signs  must  be  like, 
and  the  sum  being  negative,  one  at  least  must  be  negative. 
The  fornf  is 

{x  —  a){x  —  b)  =  x"^  —  (a  -\-  h)x  -{-ab. 

EXERCISES. 


Factor: 

1.  x""  -  bx-{-  6. 

Ans. 

(X- 

2)(^  -  3). 

2.  x"  -  ix-^  3. 

3. 

x^-  ^x^  12. 

4.  x"  -^x-{-  18. 

5. 

m*  -  5m'  +  4. 

6.  b'  -Qb'  -\-  8. 

7. 

c'  -  ^&  +20. 

8..  ^x'  -  %nx  +  60. 

9. 

5^^^  -  25^  +  20. 

10.   W  -  %W  +  21. 

11. 

-2 h  35. 

a        a 

12.   ^'\  -  11^-  +  24. 
n             n 

13. 

i'-\+- 

134.  If  the  last  term  of  the  trinomial  is  negative,  one 
of  the  quantities  a  and  Z*  must  be  negative  and  the  other 
positive,  and  the  coefficient  of  the  second  term  must  be  their 
algebraic  sum. 


DIVmON  INTO  PRIME  FAOTORS.  113 

Hence,  in  this  case,  we  must  find  two  numbers  whose 
product  is  the  third  term  and  whose  numerical  difference  is 
the  coefficient  of  a-. 

Then  prefixing  to  the  greater  number  the  sign  of  the 
coefficient  of  x,  and  to  the  lesser  number  the  opposite  sign, 
we  have  the  quantities  to  be  added  to  x  to  form  the  factors. 

EXERCISES. 

Factor: 

1.  x'  -\-^x-  8.  Ans.   {x  -\-4:){x-  2). 

2.  x'  -2x-  8.  3.  a'  -^U-  3. 

4.  a'  -^a-  3.  6.  m'  —  m  -  9. 

6.  r>i*  -{-  m  —  Q. 

7.  5a;'  -  lOx  -  15.  Ans.  ^{x  -  d){x  +  1). 

8.  5^^^  +  10:c  -  15.  9.    -[  +  2-  -  3. 

W'     '        7i 

1 3  5 .  By  combining  the  above  forms  others  may  be  found . 
For  example,  the  factors 

(a'  -{-ab-{-  h'){a'  -  ab -}-  h')  (1) 

are  respectively  the  sum  and  difference  of  the  quantities 

a^  +  ^'     a^d     ^^• 
Hence  the  product  (1)  is  equal  to  the  difference  of  the 
squares  of  these  quantities,  or  to 

(a'  +  by  -  a'b'  =  a*-\-  cfb'  +  b\ 
Hence  the  latter  quantity  can  be  factored  as  follows: 
a'  4-  a'b'  +  b*  =  {a'  +  ab  +  b'){a'  -  ab -\-  b'), 

EXERCISES. 

Factor: 
1.  m*  +  m'7i'  +  n\ 

3.  a*  +  9a'b'  +  81b\ 

5.  3x*  +  12a'x'  +  A8a\ 

7.  16  +  4rt'  +  a\ 

9.  ^:+i+-t 

a*      a'm''      m*_ 

C  C  71  71 


2. 

a'  4-  la'b'  +  Wb\ 

4. 

Ua'  +  4.a''b''  +  b\ 

6. 

U*  +  18^''^  +  162. 

8. 

?+4+''- 

10. 

^^■+>- 

12. 

i'+.^- 

Teachers  requiring  additional  exercises  in  factoring  will  find  them  in  the 
Appendix. 


114     OPEnATlONS  WITH  COMPOUND  EXPTtE88I0N8.. 


Factors  of  Binomials. 

136.  Let  us  multiply 

OPERATION. 

X  —  a 

a;"  _^  ^^  »  -  1  _|_  ^,2^  n  -  2  _^  ^^3^n-3_|_    .    _    ,    _|_  «  "  -  l^g 

—  ax""-^  —  (ex""-^—  a'x''-^—  ....  —  n^'-'^x—  a'' 
Prod. ,  a?       0  0  0  0^~-^<? 

The  intermediate  terms  all  cancel  each  other  in  the  pro- 
duct, leaving  only  the  two  extreme  terms. 

The  product  of  the  multiplicand  hy  x  —  a  is  therefore 
x^  —  a^.  Hence  if  we  divide  x""  —  a^  by  x  —  a,  the  quotient 
will  be  the  above  expression. 

Hence  the  binomial  x^  —  a"  may  be  factored  as  follows; 

Therefore  we  have: 

Theokem.  The  difference  of  equal  poivers  of  tioo  Clum- 
bers is  divisible  hy  the  difference  of  the  numbers  themselves. 

Illustration.     The  difference  between  any  power  of  5  and 
the  same  power  of  2  i»  divisible  by  5  —2  =  3.     For  instance^ 
5^-2'^:=    21  =  3.7. 
5^-2^  =  117  =  3.39. 
5*  _  2*  =  609  =  3.203. 

EXERCISES. 

Divide  the  following  expressions  \)j  x  —  a,  and  show  that 
the  quotients  correspond  to  the  above  form : 

1.  x"  -  a\         2.  x'  -  a\         3.  x'  -  a\  4.  x'  -  a\ 

Factor: 

5.  x'  -  ^a\  6.  8m'  -  ^ln\  7.  x'  -  a'x. 

137.  To  find  the  binomials  of  which  x  -]-  a  are  factors, 
let  us  call  b  the  negative  of  a,  so  that 

b  —  —  a     and     a  —  —  b. 
Then  x-\-  a  =  x  —  b. 


LOWE!ST  COMMON  MULTIPLE.  115 

Xow,  by  the  preceding  section,  x  —  <^  is  a  factor  of  a;"  —  Z>". 
Putting  —  a  in  j)lacc  of  b,  and  supposing  n  —  2,  3,  etc.,  in 
succession,  we  have 

X   —  b   =  X  -\-  a    (because  b   —  —  a), 
x^  —  b^  =  x""  —  (f  (because  b''  —  a"), 
x^  —  b'^  =  x^  -\-  CI?  (because  b""  —  —  <i^), 
x^  —  b'^  ~  x"  —  ii^  (because  b"  —  cC). 
etc.  etc. 

We  see  that  when  the  common  exponent  n  of  x  and  of  a 
is  2,  4,  G,  etc.,  a"  is  negative,  and  when  odd  it  is  positive. 
Because  all  the  above  expressions  are  divisible  by  x  —  b,  that 
IS  by  2;  -j-  ^>  we  conclude: 

Theorem.  When  n  is  odd,  the  binomial  x""  +  «"  is  divis- 
ible by  X  -\-  a. 

When  n  is  even,  x^  —  «"  is  divisible  by  x  +  ct. 

EXERCISES. 

Divide  the  following  expressions  by  x  -{-  a,  and  thus  find 
their  factors: 

1.  x'  +  a\         2.  x'  -  a\         3.  x'  +  a\         4.  x'  -  a\ 

When  n  is  even,  x^  —  «"can,  by  §  136,  be  divided  hy  x  —  a 
as  well  as  by  :r  +  ^«  Therefore  both  of  these  quantities  are 
factors,  and  the  binomial  may  be  divided  by  their  product, 

x""  —  a". 

EXERCISES. 

Divide  the  following  by  x^  —  a^,  and  thus  factor  them: 
1.  x'  -  a\         2.  x'  -  a\         3.  x'  -  a\         4.  x''  -  a'\ 


Lowest  Common  Multiple. 

138.  The  L.C.M.  of  any  polynomials  maybe  found  by 
factoring  them  and  applying  the  rule  of  §  55. 
Example.     Find  the  L.C.M.  of 

^ct"  -  2b\     a'  +  2ab  +  b%     a'  -  2ab  +  b\     3a*  -  3b\ 
Factoring,  we  find  these  four  expressions  to  be 
^a  -f  b){a  -  b),  (a  +  b)',  {a  -  b)\  3{a'  +  b'){a  +  b){a  -  b). 
Applvmg  the  rule,  avo  find  the  L.C  M.  to  be 
2  .  3(a  -[-  by{a  -  bfia'  -f  b'). 


116    OPERATIONS  WITH  COMPOUND  EXPRESSIONS. 

EXERCISES. 

Find  the  L. CM.  of: 

1.  a'  -  b\  a'  -  b\  2a  +  2b. 

2.  a'-l,  a'  -]-a-2,  a-1,  a-\-  2. 

3.  2a  -  1,  4a*'  -  1,  4ta'  +  1. 

4.  x'  -  X,  x:'  -1,  x'  -^1,  x-\- 1. 
b.  X,  X  —  a,  x^  —  a^f  X  -\-  a, 

6.  X  —  a,  x^  —  ax,  x^  —  ax"^. 

7.  iy'  -  W,  4:y'  +  U%  2y  +  2b. 

8.  x'  —  a\  x^  —  a\ 

9.  Cy  a,  a  —  6',  a-\-  c. 

10.  21a'  -  ^c\  9a*^  -  U\  da  +  2c. 

11.  2a  -  b,  4:a'  -  b\  4a*  +  b\ 

12.  a^  4-  4a^  +  46%  a^  -  ^ab  +  46=^,  a»  -  4^>'. 

13.  %  +  z),  y{x  -  z),  xyz. 

14.  m  —  n,  m^  —  n",  m'  —  w^ 

15.  7?i  —  n,  m^  —  2mn  -\-  n^,  m^  —  n\ 

16.  a-\-b,  a'  -  b\  a'  +  b\ 

17.  46-'  -  ^cn  +  7i%  4c2  -  n". 

18.  a  +  1,  a''  +  1,  a^  -  1. 

19.  a;  —  4,  a;''  -  16,  a;'  -  8,  a;  -  2. 

20.  a  —  6,  b  —  a,  a-{-b,  b  -\-  a. 

21.  a:  -  «/,  ?/»  -  ic%  x-\-  y. 

22.  it?  -  2^,  ^  -  2p. 

23.  ^>*  —  a^c%  ac  —  b,  a'^c'  —  2abc  +  b\ 

24.  aa:  +  ay,  x  -\-  y,  a. 

25.  ex  +  c?/,  cV  -  6'Y,  c'. 

26.  2fla:,  3a'^a:%  4:a'x\ 


Sectiot^^  y.    Fractions. 

139.  Fractions  having  compound  expressions  for  the 
numerators  or  denominators  may  be  aggregated,  multiphed, 
divided  or  reduced  by  the  methods  of  §§  57  to  76.  The 
general  rule  to  be  followed  is: 

1.  Observe  what  part  of  the  expression  constitutes  the 
numerator,  and  what  the  denominator. 

2.  Operate  o?i  these  expressions  accordmrj  to  the  rules  prcr 
scribed  for  fract\on9% 


FRACTIONS.  117 

EXERCISES. 

140.  Multiplicatio7i  by  Entire  Quantities,  Execute  the 
following  multiplications: 

^    a-\-  h       '         , ,         .        a^  —  b"^ 

1.  — - —  Xia  —  b).       Alls. . 

x-y  x-y 

2.  ^-±^-  X  (^  +  2b),  3.  ?^  X  {2x  -  y). 
x  —  y       ^           '  a-\-b       ^  ^' 

.    ax -{-by      ,  ,   .  „    1  +  a;  +  2a;^       ..        ^  . 

4.  — -V---  X  (ay  -  bx).  5.   ,  ,  ^r-,  X  (1  -  2iz:). 

„    d^—ab-\-¥       ,        ,,  ^     a^b       ,     ,     ^ 

'■  ^+I~  X  ("  -  *)•       ^-  ?^^  X  ("  +  2')- 

Here,  because  x^  —  y^  =  (x  —  y){x  +  ^),  the  multiplier 

is  a  factor  of  the  denominator;  so  we  operate  by  §  60,  getting 

a-\-  b       ,, 

as  the  answer. 

x-y 

8-  :^4? ^  (^ - ^^)-        '•*•  ;7^^+A^ ><  (» - *)■ 

Here  denominator  and  multiplier  have  the  common  fac- 
tor a  —  2x,  which  we  suppose  cancelled.  Multiplying  by  the 
remaining  factor,  we  get 

{a-^x){a-{-  2x)  _  a'-^  Sax  +  2x'' 

-  — _T •.     Ans. 

a  —  2x  a  —  Zx 

11.      ,  ^^  '-—-.  X  K-^O-  1^.   -T^A  X  {a-\-b).  ■ 

7ri  —2m7i-\-n^      ^  ^  a  -\- b       ^     ^     ^ 

fl^  +  2^  a  — 2b      ,        -.  * 

13.  ^ri:^iX(a+M.  14.   -^^---,x(a-b). 


1^-  j-r^-T^i  X  (^  + .).     16.  ^^:^^:^^  X  (b  -  c). 
i^-^xK-5^). 

Here  the  denominator  is  a  factor  of  the  multiplier,  so  thq 
product  is  an  entire  quantity,  namely, 

a  X  (a  —  b)  =  a""  —  ab. 

18.   ^^-  X  {m'-  2}nn^n').  10.   "-^\  x  ia'  -  b'). 


118     OPEltATIOSS    WirU  COMPOUND  EXPUliti^IOA-S. 

3S.   i!i+|^  X  (..'  -  8/).  23.   ^iil^-  X  (..  -  V). 

x  —  'Zy      ^  ^  '  ax  — ay      ^        '' ' 

141.  Dividing  hy  dividing  the  Numerator  or  7nuUi]jly- 
ing  the  De7iominator  (§§  58,  59). 

a  -\-b         ^         ^' 


Ans. 


4. 

a-b      ''    ("+^)- 

5. 

.+  1^(^+^)- 

6. 

xXx''^^       !)• 

7. 

.-2-^    ^(^  +  ^^)- 

1 

x'  -\-"Zx-{-X 


8.  ^4-K-  ^')- 
a  —  0 

9.  — ^  4-  (a  -  2^).  10.  -^^--  -^  (?yi^  -  ^r). 

11.  ^^-  -  (2^  +  c).  12.  -^^|-  -^  (7/i  +  n). 

b  -{-2c  m  —  2n       ^      '      / 

13.   5£_^^^  _^  (:^;«  _  y^).  14  — ^  -  (a.-^  -  f). 

X  -\-  y         ^  ^  '  x  —  y        ^  ^  ' 

15.   -^  "^  '\     ,  --  (2;^  -  ?/^). 


142.  Reduction  to  Lowest  Terms  (§65). 

2. 


..  2.      „        — „ 

ic   —  rt 

X  —  a 


a;^ 

~r 

3. 

X 

-\-a 

x' 

+  «" 

^ 

x' 

-  2^y + 

// 

t). 

^^-2/^ 

x^  —  a' 

a  —  b  -\-  c 


(f  -  2ab  +  b'  -(f 


FRACTIONS.  11^ 

am  +  hn  +  ^mx  h^  +  GZ>g  +  9^' 

a'x'  -  2abxy  +  by  a' -  2a'b -\- a'b' 

^'       '^-by      *  «^-a'^^'^    • 

c'-[-^c^x-^4:d'x''  1  —  X 

~^V^=^1^^''~'  1  -  o;^* 

bx  +  by  '    «'—  a^  +  b^' 


^2_  5;;^  +  6*  *  x'-x-  6* 

143.  Reduction  to  Given  Denominator  (§67). 
1.  Express  a  -{-  b  with  denominator  a  -^  b. 

(a^bY       a'  +  2ab-}-b' 
Ans.    ^     '     ^   — 


a  -{-  b  a  -\r  b 

with  denomi 


2.  Express  ,     with  denominator  a^  —  J^ 

3.  Express 


a-{-  2x 

4.  Express  7-  " 

5.  Express  —  ** 

6.  Express—^—     " 

7.  Express  ^^ 

X  —  i 

8.  Express      ,^  ^        '' 

^        a  -^  2x 

9.  Express 


a" 

—  ^x\ 

J:r 

+  %. 

W^2 

^  +  ^^ 

m^ 

-n\ 

a;* 

-1. 

46? 

»^'  -  «*. 

x' 

-a\ 

a  —  X 

10.  Express  — .   ^   . —  with  den.  <5^'  +  2ac  —  c'  +  J'» 

a-\-b  -\-  c 

11.  Express  — i-r  ''     *'     ^^  -  5a;  +  6. 

■^        a;  —  2 

12.  Express  ^\  '^  ^\'\  1'  "     "     a*  -  ^'^'^  +  ^*. 

13.  Express ^ "     "     m'  -  2w/^  +  w'  -  a;'. 

m  ^  n  —  X 


120     OPERATIONS  WITH  COMPOUND  EXPRESSIONS. 

144,  Reduction  to  L.C.D.  and  Aggregation.  The  fol- 
lowing algebraic  sums  of  fractions  are  to  be  reduced  to  their 
L.C.D.,  and  then  aggregated  into  single  fractions  by  the 
methods  of  §§  68  to  72. 

a-b^a-\-h'  a'-¥~~  a' -b'' 


a  -{-b      a  —  b' 

3    ^  +  2y       x-^  ^    1_ 1__ 

'  X  —  %y       x-\-%y'  'a       a  ^  b' 

K    _i L_  f.        a       ,       ^ 


a-\-b      a  —  b'  '  h  —  c~  c  —  b' 

1.1.1  ^        a  a'       .       a' 


9. 


a    '   a-[-b   '   a  —  b'  'a  —  b      a""  —  b""  ^  a' —  b*' 

a^b      a-b  AF 


a-b      a-\-b      a^ -¥' 

10.  — — '—-.  11.  a  +  -7 . 

i?;+la;  —  1  b  —  a 

11  n^ 

12.     1    -  :; :r-^^.  13.  a  -\-  b ^ 

1  —  a       1  —  a  a  —  b 


14.  «  +  i  +  -_^.  15.  ;.  -  2,  - -^^i^. 

16.       ^  ^ 


1 - c      1  +  c 

1  1 


4- 

10c 

c''- 

-1  * 
3 

^'^'  a-l      o  +  2      a'+  ia  +  4' 

18.  i±l-^^_(^  +  y)  1  +  1). 

19.  (w  +  7i) —  (m  —  ^i) . 

^  ^  \n       m)       ^  '  \n       m  j 

20.  (a-*)(i-i)  +  («  +  J)(l+y. 

33.  5^-+-!-  +  ^.  24.  i  +  ^'-*^y+%-. 


FRACTIONS.  121 

145.  MuUipUcation  of  one  Fraction  hy  Another  (§  75). 

Note.     In  these  exercises,  see  whether  common  factors 

can  be  cancelled  before  multiplying. 

a         x'  -¥ 
X 


'  X  —  b  b 

We  see  that  the  first  denominator  and  the  second  numerator  have 
the  common  factor  x  —  b.     So  we  cancel  it,  and  only  multiply 
x  +  b_a{x  +  b) 

''^~~r~     b    ' 

2     ^  +  ^    V   ^  +  y  Ans    L 

^-  ^-^-.y^^  a"  -  b''  ^'  (a  -b){x-'  y)' 

„    a  -\-b  —  c      a  —  b  —  c 


a  -{-b  -\-  c      a  —  b  -\-  c 

-G-??)(|-^)('  +  r^)- 

^    a^b  I  a'      ,       b' 

a  —  b  \        a  — 


b   '  a-\-b. 


6.   ^^f-l+       -^ 


x-\-  Qy  \  X  —  ^y      X  -{-  2yJ 

^    a^  -\-ab        a^  —  b^ 


a'  +  ^'       «'^  +  «^'' 

_    «'  —  '^ax  -\-  x"^          X  a^  A-  X* 

8.   . ! X X  — -T— . 

a  -\-  X             a  —  X  X 

1      «-  +  y   «+a 

a  —  b        a  —  b       a  —  b 

-•^    ^  +  V      x"^  —  y^  1 

10.   — ^-^  X    3   ,      3  X 


11    _^lzZ_v    ^-^ 


a'-Ub^b'''    a'-\-ah' 

12.    fl  +  ^  +  ^Vi-!i  +  ^). 
V        X    '   aj\        X    ^   aJ 

146.  Division  by  inverti7ig  the  Divisor  (§  76). 
The  dividend  and  divisor  are  first  to  be  aggregated  and 
reduced  if  necessary, 

1.  n ^-^  n . 

^  n 

n*-l  ^  n^—l  _  n'-l  n      _  n^-\-  1 

n  n  n  n  —  1  n 


122     OPERATIONS  WITH  COMPOUND  EXPRESSIONS. 

a       ^       h          a  —  h 
^    _j — _  _i_ ^ 

'  a  -{-  b       a  —  b       a-\-b 


71  —  1  71  -\-  1  71  -{-1 

'  a-\-b       a  —  b  '    a-\-b' 

a  X       ^      2x 

X  -\-  a      X  —  a   '  X  —  a 

a  -\-b       a  —  b  \      2a 

'  a  —  b      a  -{-  b  '   a  -\-  b' 

9,       '  '  '  ' 


x-1       x-^1       x'   '   x\x'  -  !)• 

a  -}-  b       m  —  n   ^  1 

a  —  b       7n-{-  71   '    {a  —  b)  {in  -\-  71)' 

13.  l  +  ^!_  +  ^+    •   . 

a  —  X       a  —  X        X  -f-  a 

147.  deduction  of  Complex  Fractions. 

Def.  A  complex  fraction  is  one  either  or  both  of 
"vrhose  terms  is  fractional. 

The  minor  fractions  are  those  which  enter  into  the 
terms  of  the  complex  fraction.  Their  terms  are  called 
minor  terms. 

Problem.     To  reduce  a  complex  fraction  to  a  simple  one. 

EuLE.  Multiply  both  terms  by  the  L.C.M.  of  the  minor 
denominators. 

Reason.     1.  The  value  of  the  fraction  remains  unchanged 

(§64)- 

2.  The  minor  denominators  are  removed. 
m       n 

Example  1.     Reduce to  a  simple  fraction. 

m       n 

n       7n 


FRACTIONS.  123 

The  minor  denominators  are  m  and  n,  and  their  L.C.M. 
is  fan.     Then: 

Numerator,      1 X  mn  =  m^  +  n*. 

71       m 

Denominator, X  mn  =  m^  —  n^. 

n       m 

Therefore  the  given  fraction  is  equal  to  —2—^^ — ^. 


m 


m       m 

—       —Xnq 

Ex.  2. 


n  _  n         "  _  mq 

p  ~  p  ^  np' 

--      —  X  nq        ^ 
q        q 


EXERCISES. 

Beduce  the  following  complex  fractions  to  simple  ones: 


m 

3.  — -% 

m 
m 

y 

a 

5.  J. 


7. 


9. 1 


n  —  \ 

^-  1 
1 


1  +  - 


n.  ^ 


1  - 


9 

'+r 

« 

1  — ^ 

h 

a 

b 

4. 

X 

y 

a-\-  b 

a-b 

6. 

a  —  b' 

a-\-  b      • 

a              b 

8. 

a+b ' a-b 

a              a 

a-\-b  '   a—b 

10. 

'      1. 

a 

1 

X 

X 

1+1-. 

VZ. 

1  +  x     • 

124     0PEUATI0N8  WITH  COMPOUND  EXPBESISIONS, 

1  1  1 

14.   ~ . 


13. 

-      l-x 

X 

15. 

^      1  +  x 

m  —  n 

1   1   ^-^ 

17. 

10 

'   m  +  n 

-  +  -+- 
x^y^  z 

x^  y  '^  z 

1 

x-\-y 

X  -\-  a  X  —  a 

X  —  a  X  -\-  a 

X  -\-  a  X  —  a 

X  —  a  X  -\-  a 
a              b 


18.  "-*      "+^. 
a  0 

a  -\-  b      a  —  b 

m       m  —  n 

111-  20.   KZ»±». 

— h  T  H —  1^'    ,   M  4-  n 

a       b       c ' — 

71       m  —  n 


Section  YI.    Substitutioi^. 

148.  In  algebra  we  often  have  to  substitute  an  expres- 
sion for  a  single  letter  in  some  other  expression.  This  is 
done  by  making  the  substitution  as  in  Chapter  II.,  §  109,  and 
then  reducing. 

EXERCISES. 

Make  the  following  substitutions: 

1.  In  a;^  +  -5-  substitute  x  =  -,  Ans.  --  4-  cC 

X  a  a^  ^ 

2.  In substitute  x  ='  —. 

1  —  X  a 

X 

3.  In  :; substitute  x  =  1  —  a. 

1  —  X 

4.  In  -— —  substitute  x  = 


l-x  I-  a 

5.  In  1 substitute  x  :=  1 . 

X  a 

G.  In  x^  +  2  +  -2-  substitute  x  = y. 


THE  G.G.D.  OF  TWO  J^UMBEM8,  125 

7.  In  ax"  -j-  Ifx  substitute  x  =  a  —  I), 

8.  In  -n- substitute  x  =  -r, 

X         a  0 

0^        x^ 

9.  In  — substitute  x  =^  a, 

x        a 

ft      I      QC 

10.  In  substitute  x  =  am, 

a  —  X 

11.  In  — ■ —  substitute  x  = . 

X  -{-  a  a  —  c 

12.  lix  = T  and  b  = ,  find  x  in  terms  of  c, 

1  —  b  1  —  c 


Section  VII.     The  Highest  Common  Divisor. 

The  G.C.D.  of  Two  Numbers. 

149.  Theorem  I.  If  ttvo  numbers  have  a  common 
divisor,  their  su7n  will  have  that  same  divisor. 

Proof,  Let  <^  be  a  common  divisor; 

m  the  quotient  of  one  number  divided  by  r/; 
n  the  quotient  of  the  other  number  divided 
byrf. 
Then  the  two  numbers  will  be 

dm    and    dn\ 
and  their  sum  is  d{rn  +  n). 

This  sum  is  evidently  divisible  by  ^;  and  the  quotient 
w  +  w  is  a  whole  number  because  m  and  n  are  whole  num- 
bers; hence  follows  the  theorem  as  enunciated. 

Theorem  II.  If  two  numbers  have  a  common  dirisor, 
their  difference  will  have  that  same  divisor. 

Proof.     Almost  the  same  as  in  the  last  theorem. 

Cor.  If  one  number  divides  another  exactly,  it  will  divide 
all  multiples  of  that  other  exactly. 

Remark.  The  preceding  theorems  may  be  expressed  as 
follows: 

A  common  divisor  of  two  numbers  is  a  common  divisor  of 
their  sum,  difference  and  multiples. 


126     OPERATIONS  WITH  COMPOUND  EXPRESSIONS. 

Eemark.  If  one  number  is  not  exactly  divisible  by 
another,  a  remainder  less  than  the  divisor  will  be  left  over. 
If  we  put 

D  =  the  dividend; 
d  =  the  divisor; 
q  =  the  quotient; 
r  =  the  remainder; 
we  shall  have  I)  =  dq  -\-  r, 

or  D  —  dq  =  r. 

Example.  7  goes  into  %Q  9  times  and  3  over.  Hence 
this  means 

66  =  7  .  9  +  3,     or     66  -  7  .  9  =  3. 

150.  Pkoblem.  To  find  the  greatest  common  divisor  of 
two  numbers. 

Let  J[f  and  N  be  any  two  numbers,  and  let  if  be  the  greater. 

1.  Divide  if  by  N.  If  the  remainder  is  zero,  iV^will  be  the 
common  divisor  required,  because  every  number  divides  itself. 
If  there  is  a  remainder,  let  q  be  the  quotient  and  R  the 
remainder. 

Then  M  -  Nq  =  R. 

Let  d  be  the  common  divisor  required. 

Because  M  and  iV^  are  each  divisible  hj  d,  M  —  Nq  must 
also  be  divisible  by  d  (Theorem  II.).     Therefore 
R  is  divisible  by  d. 

Hence: 

a.  Every  common  divisor  of  M  and  N  is  also  a  common 
divisor  of  JV^and  R. 

Conversely,  because 

M=  Nq-{-  R, 

/?.  Every  common  divisor  of  iV^and  R  is  also  a  divisor  of 
M,  and  therefore  a  common  divisor  of  ilf  and  Nf. 

Comparing  a  and  ^  we  see  that  whatever  common  divisors 
M  and  iV^  may  have,  those  same  common  divisors  and  no 
others  have  aYand  R. 

Therefore  the  greatest  common  divisor  of  M  and  N  is  the 
same  as  the  greatest  common  divisor  of  N  and  R,  and  we 
proceed  with  these  last  two  numbers  as  we  did  with  Jf  and  N, 

2.  Let  R  go  into  N q'  limes  with  the  remainder  R\ 
Then  X  =  Rq'  +  R', 

or  NT  -  Ra'  =  J?\ 


THE  Q.a.D.    OF  TWO  NUMBERS.  127 

Then  it  can  be  shown,  as  before,  that  d  is  a  divisor  of  R'y 
and  therefore  the  greatest  common  divisor  of  R  and  R'. 

3.  Dividnig  R  by  R\  and  continuing  the  process,  one  of 
two  results  must  follow.     Either — 

a.  We  at  length  reach  a  remainder  1,  in  which  case  the 
two  numbers  are  prime  to  each  other;  or, 

/?.  AYe  have  a  remainder  which  exactly  divides  the  pre-' 
ceding  divisor,  in  which  case  this  remainder  is  the  common 
divisor  required. 

To  clearly  exhibit  the  process,  we  express  the  numbers 
J/,  i\^and  the  successive  remainders  in  the  following  form: 


M  = 

JV.  q  +  R,                    (R     <  J^); 

jsr  = 

R.q^-i-R\                     {R'     <R)', 

R  = 

R\q''-^R'',                    (i?"    <R')) 

R'  = 

R".  q'"  +  i^'",             (R'"  <  R''); 

etc. 

etc.                                    etc.. 

itil  we  reach  a  remainder  equal  to  1  or  0,  when  the  series 

rminates. 

Example. 

Find  the  a.C.D.  of  240  and  155. 

240  -^ 

155  gives  quotient  1  and  remainder  85 

155  ~ 

85  gives  quotient  1  and  remainder   70 

85  -V- 

70  gives  quotient  1  and  remainder  15 

70  ~ 

15  gives  quotient  4  and  remainder  10 

15  ^ 

10  gives  quotient  1  and  remainder     5 

10  -^ 

5  gives  quotient  2  and  remainder     0. 

Therefore  5  is  the  greatest  common  divisor. 

EXERCISES. 

Find  the  G.C.D.  of  the  following  pairs  of  numbers,  ar- 
ranging the  work  as  in  the  above  example: 

1.  12  and    56.  2.  232  and    144. 

3.   96  and  156.  4.   108  and    153. 

5.   72  and  102.  6.   158  and  1024. 

151.  Case  of  Three  or  more  Numhers.  To  find  the 
(t.O.D.  of  three  or  more  numbers,  we  first  find  that  of  any 
two,  and  then  the  G.C.D.  of  this  G.C.D.  and  the  third 
number. 

EXERCISES. 

Find  the  G.C.D.  of: 
1.   12,  15,  39.  2.   98,  140,  217.  3.   270,  198,  153. 


128     OPERATIONS  WITH  COMPOUND  EXPRESSIONS. 


The  H.C.D.  of  Two  Polynomials. 

15^,  The  theorems  of  §149  relating  to  two  numbers 
apply  equally  to  two  polynomials.  Hence  we  may  find  the 
H.C.D.  of  two  polynomials  by  a  process  similar  to  that  of 
finding  the  G.C.D.  of  two  numbers. 

Example.     Find  the  H.C.D.  of 

x'  -2x'  -x'  -\-x-i-l 
and  X*  -}-  x^  —  X  —  1. 

First  Division. 

x'  -  2x'  —    x'  -{-    x-i-l  Ix^-^-x'  —  x  —  l 

X  -  1 


x'-\-x' 

-    x'  -    X 

-x*  -  2x' 
-X*  -     x' 

+  2:r+l 

4-   ^  +  1 

-^    X  =  first  remainder. 


Second  Division. 

X*  -\-  X^  —  X  —  1  I  —  x^  -\-x 
X*  —  x^  —  X   —  1 


x^  -]-  x"^  —  X  —  1 

X^  —  X 


1  =  second  remainder. 


Third  Division. 

-x'  ^x  I  x'  -  1 

—  X^  -\-  X  —  X 


0        0     =  third  remainder. 
Thus  a;'  -  1  is  the  H.C.D.  sought. 

EXERCISES. 

rind  the  H.C.D.  of: 

1.  x"  ~1  and  a;"  —  1.  2.  x^  —  a"  and  x*  —  a\ 

3.  rr'  -  6a;  +  8  and  4a;'  -  21a;'  +  15a;  +  20. 

4.  x"  +  a;'  and  a;*  —  1. 

153.  Case  of  Factors  found  by  Sight.  When  a  common 
factor  of  the  two  polynomials  can  be  found  by  simple  in- 
spection, we  remove  this  factor  from  both  polynomials,  find 
the  H.C.D.  of  the  quotients  and  multiply  it  by  the  factor. 


THE  H.G.D.  OF  TWO  POLYNOMIALS.  129 

Example.     The  polynomials 

x'-\-x    and     x' -{- x' ^  x' -\- x"  (1) 

have  the  common  factor  x.  This  factor  is  therefore  a  factor 
of  the  H.O.D.  sought.  Now  if  we  divide  it  out  the  polyno- 
mials will  become 

x'-\-l     and     x' ^x' -^x" -\-x.  (2> 

If  now  we  find  the  H.C!D.  of  these  expressions  (2)  and 
call  it  D,  then  Dx  will  be  the  H.C.D.  of  the  given  polyno- 
mials (1).     We  shall  find,  by  going  through  the  process, 

D  =  x^l. 
Therefore  the  H.C.D.  of  (1)  is 

x""  -\-  X. 

154.  Throwing  out  Factors.  If  we  carry  on  the  pre- 
ceding process  without  modification,  we  shall  commonly  find 
that  numerical  fractions  enter  into  the  remainders.  These 
may  be  avoided  by  applying  the  following  principle: 

If  a  divisor  contains  any  factor  prime  to  the  dividend,  if 
may  he  rejected  before  dividing. 

The  reason  of  this  is  that  we  are  seeking,  in  the  final  re- 
sult, only  for  the  product  of  all  those  factors  which  are 
common  to  both  divisor  and  dividend.  Therefore  a  factor 
contained  in  one  but  not  in  the  other  is  not  a  factor  of  the 
H,C.D.  sought,  and  hence  may  be  rejected. 

For  a  similar  reason,  we  may  multiply  any  divide7id  by 
any  factor  prime  to  the  divisor. 

Example.     Find  the  H.C.D.  of 

x'  -  4:x'  +  12a;'  +  4:x'  —  13:r 
and  X*  -  2x'  +    4:x'  -{- 2x  -    5. 

First  Division. 

x'  -  4.x*  -f-  12a;''  4-  4:x'  -  13a:  |  x*  -  2x'  -}-  4a;'  +  2a:  -  5 

x'  -  2a:^  +    4:x'  +  2a:'  -    5x    x  -2 

_  2x'~-{-    Sx'  +  2a;'  -    8a: 

-  2a;*  -j-    4a:'  -  8a;'  -    4a;  +  10 


4a;'  +  10a:'  —    4a;  —  10  =  first  remainder. 

This  remainder  contains  the  factor  2,  which  is  not  con- 
tained in  the  dividend.  So  we  divide  by  it.  But  then  the 
first  term  of  the  next  divisor,  2a:',  will  still  not  go  into  ar* 


130     OPERATIONS  WITH  COMPOUND  EXPRESSIONS. 

without  a  fractional  quotient.     So  we  multiply  the  new  divi- 
-dend  by  2. 

Second  Division. 

'^x'  -  4.x'  +    ^x""  +  4:x  -  10  I  2x'  -\-5x'  -2x-6 


2x'  +  5:^:='  -    2x'  -  5x  ^'  -  | 

-  9x'  +  10^:'  +  \^x  —  10 

-  dx'  -  ^x'  H-  9^  +  -¥ 

-^^x""  —  -%^-  =  second  remainder, 

or  \H^^  ~"  1)  =  second  remainder. 

To  have  avoided  all  fractions,  we  should  have  multiplied 
the  dividend  by  4.  But  we  could  not  know  this  until  after 
we  had  begun  the  division,  and  the  failure  to  multiply  does 
no  harm. 

We  now  reject  the  factor  %S-  from  the  remainder,  leaving 
ic^  —  1  as  the  next  divisor. 

Third  Division. 

2x'  -\-Dx'  -2x-  d\    x'  -1 


2x^  —  2x  2x  +5 

5x'  -  5 
6x'  -  5 

0        0  =  third  remainder. 
Hence  ^^  —  1  is  the  H.C.D.  sought. 

EXERCISES. 

Pind  the  H.C.D.  of: 

1.  x'  +  x'   and   x'  -  2x'  +  2x'  -  2a;  +  1. 

2.  2x'  +  x'  -  5a;  +  2   and   4:x'  -  ^x'  -  5a;  +  3. 

3.  a;'  +  1   and   x'  +  ax""  -\- ax -{- 1. 

4.  x'  -  ^x'  +  2a;'  +  a;  -  1  and   x'  -  x'  -  2x  +  2. 

5.  2a;'  -  llo;'  -  9  and   4a;'  +  11a;'  +  81. 

6.  6x'  -  Hax^  -  20a'x  and   3a;'^  +  «a;  -  4«l 

7.  12a;*  +  4a;^  +  17a;'  -  3a;  and   24a;'  -  52a;^  +  14a;' 

8.  a*  -  a' -{-20"  -\- a -\- 3   and   a'  -{- 2a'  -  a  —  2, 

9.  ea*  +  a'  —  «  and   4:a'  —  6a'  —  4«  +  3. 


CHAPTER  III. 

EQUATIONS  OF  THE  FIRST  DEGREE. 


Section  I.    Equations  with  One  Unknown- 
Quantity. 

155.  Bef.  An  equation  of  the  first  degree  is  one 
which,  when  cleared  of  fractions,  contains  only  the 
first  power  of  the  unknown  quantity. 

All  equations  of  the  first  degree  may  be  solved  by  the  pro- 
cesses of  multiplication,  transposition  and  division  explained 
in  §§  82  to  86.  These  processes  are  embodied  in  the  follow^ 
ing 

KuLE.      1.   Clear  the  equation  of  fractions, 

2.  Transpose  the  terms  which  are  multiplied  hy  the  un- 
known quantity  to  one  meml)er  and  those  which  do  not  con- 
tain it  to  the  other. 

3.  Aggregate  the  coefficients  of  the  unknown  quantity;  and 

4.  Divide  loth  members  by  the  coefficient  of  the  unknown 
quantity. 

Example  1.     Let  us  take  the  equation 
x  —  1    _  2a;  —  6 
2^  +•  10  ~"  Ix  +2* 
Clearing  of  fractions,  we  have 

^x"  -  ^Qx  -  14  =  ^x""  +  8a;  -  60. 
Transposing  and  reducing, 

46  =  34a;. 
Dividing  both  members  by  34, 

_46_23 
^  ~  34  ~  17* 

This  result  should  now  be  proved  by  computing  the  values  of  both 

23 
members  of  the  original  equation  when  —  is  substituted  for  x. 


132  EQUATIONS  OF  THE  FIB8T  DEGSEE. 


Ex.3.  -^ ^  =  -.-P- 


X  —  a      X  -\-  a      x^  ^  d^ 
Here  the  L.C.M.  of  the  denominators  is  7^  —  a'.     Multi- 
plying each  term  by  this  factor,  the  equation  becomes 

m{x  -\-  a)  —  n{x  —  a)  =  p^ 
or  (m  —  n)x  -{-  {m  -\-  n)a  =p. 

Transposing, 

{m  —  n)x  z=  p  —  am  —  an; 

,  p  —  am  —  an 

whence  x  =  ^ 

m  —  n 

EXERCISES. 

Solve  the  following  equations,  regarding  x,  y  or  u  as  the 
unknown  quantity: 

1.  i  +  I  =  1.  2.  -^—  +  -^  =  2. 

a       0  X  —  a      X  —  0 


u  —  d      u  —  1  u  -\-a      u  -\-^a 


^    5^-1       9t/-6^9y-7 

7       "^      11  5 

7.   {x  -2)  {x-5)  =  {x-  S)x. 

.0   ^ ?y__o  9       ^,1^6 

i/-l      2-y  2y-  6~^  y  -3      Sy  -  1' 


10. 


Note.     When  common  factors  appear,  divide  them  out. 
1111 


X  —  2      X  —  3      X  —  ^ 


X  —  a  _x  —  b  a-\-b_      a  b 

11.     ; — 7  —  ; .  1/5.     — 7. 

X  -{-  0      X  -f-  c  X  —  c      X  —  a      X  —  0 

X  —  a      X  ^  b      X  —  c  _x  —  {a  -\-  b  -i-  c) 

±0,  — — . 

0  c  a  abc 

14.  ax-\-b  =  -^\, 

a       b 

^  „    7n(x  +  a)  .  n{x  +  b)  , 

15.  — ^^ — r^-^  -\ — ^^ — r — -  =  m-\-n. 

x-\-b  X  -^  a 

16.  (x  -  a)'  +  (a;  -  b)'  +  {x  -  c)'  =  S(x  -  a){x  -  b)(x  -  c). 


EQUATIONS  WITH  ONE  UNKNOWN  qUANTITY.    133 

17.    «.Il^-^^=0.  18.  l  =  a. 

u  -\-m        u  -\-  n  X 

19.  l  =  i.  20.  l  =  i. 

X       b  X       b 

a;  —  fl5^4~^  ^       X  —  a      X  —  0 

mx      ,      nx  . 

X  —  m      X  —  71 

\yt  n't 

Find  the  value  of  -  from  each  of  the  following  equations 

X 

without  clearing  of  fractions: 

26.   -  =  4.     Ans.  -  =  2.  27.  -  =  15. 

XX  X 

38.  ^-  =  a«.  29.  ^L±_l  =  ,«. 

X  X 

30.  ■ —  =zm  —  n,  31. —  =  m'  —  rf. 

X  XX 

32.  TT-  +  o -  =  «•  33. \-—  =  a  —  b, 

2x      2x  ax      bx 

Find  -  and  z  from: 

z 

S4,    —--=z-  ^^    m-\-n  _m-\-n 


z       a       b  z  m  —  n 

36.  ^^-  =  ^.  37.     "  +  "  -     ^ 


W2      ^  mz  —  nz      m  —  n 

38.^^^  =  "-.     '  39.  ^  +  l  +  f=^. 
(a  +  c)2;      c  2        2;z      3;? 

In  the  following  equations  find  the  value  of  each  symbol 
in  terms  of  the  others: 

40.  2x-^a=zbb. 

.  '^x-bb  ^      2x  -da  3a  +  6b 

Ans.  a  =  —^-;  b  =  — ^-;       x  =  — ^— . 

41.  5a- U  =  2u,  42.   Ha  -  Ub  =  21y. 
43.   ax  =  by.  4:4:.  a(x—  y)  =  b{x  -\-  y). 

45.^  =  ^.  46.^  =  2. 
b  —  c      2a  —  c  by 


134  EQUATIONS  OF  THE  FIRST  DEGREE. 

Section  II.     Equations  of  the  First  Degree 
WITH  Two  Unknown  Quantities. 

156.  Def.  An  equation  of  the  first  degree  witli 
two  unknown  quantities  is  one  which  admits  of  being 
reduced  to  the  form 

ax-\-hy=  (?, 
in  which  x  and  y  are  the  unknown  quantities  and  a,  b 
and  c  represent  any  numbers  or  algebraic  expressions^ 
which  do  not  contain  either  of  the  unknown  quanti- 
ties. 

Def.  A  set  of  several  equations,  each  containing 
the  same  unknown  quantities,  is  called  a  system  of 
simultaneous  equations. 

157.  To  solve  two  or  more  simultaneous  equa- 
tions, it  is  necessary  to  combine  them  in  such  a  way 
as  to  form  one  equation  containing  only  one  unknown 
quantity. 

Def,  Elimination  is  the  process  of  combining 
equations  so  that  one  or  more  of  the  unknown  quanti- 
ties shall  disappear. 

The  term  "elimination"  is  used  because  the  unknown 
quantities  which  disappear  are  eliminated. 

There  are  three  methods  of  eliminating  an  unknown 
quantity  from  two  simultaneous  equations. 

Elimination  by  Comparison. 

158.  EuLE.  Solve  each  of  the  equations  with  respect  to 
one  of  the  unknown  quantities  and  put  the  tioo  values  of  the 
unknown  quantity  thus  obtained  equal  to  each  other. 

This  will  give  a  new  equation  with  only  one  unknoivn  quan- 
Hty,  of  which  the  value  can  he  found  from  the  equation. 

The  value  of  the  other  unknown  quantity  is  then  found  hy 
substitution. 

Example.     Solve  the  following  set  of  equations: 
x-^y  =  28,) 
Sx  -2y  =  29.  ) 


ELIMINATION  BY  SUBSTITUTION. 


135 


From  the  first  equation  we  find 

a;  =  28  -  «/, 
_29_+2^. 
~       "3       ' 
_  29  +  2^ 


and  from  the  second 


from  which  we  have  2S  —  y  = 

whence  i/  =  11. 

Substituting  this  talue  in  the  first  equation  in  x,  it  becomes 

a;  =  28  -  11  =  17. 

If  we  substitute  it  in  the  second,  it  becomes 

29  +  22      51       " 
^=-3— =  -3-  =  17, 

the  same  value^  thus  proving  the  correctness  of  the  work. 

EXERCISES. 

Find  the  values  of  x  and  y  from  the  following  equations: 

x-3y=    7. 


1.     x-i-2y  =  ll! 

'^,  %x—    y  =  8 

3.  X  —  2y  =  b 

4.  2x  -Sy=  7 

5.  2x  —  3y  =  7)1 
7        2 

X       y' 


3x-i-2y  =  40. 

3y  —     x=     8. 

2x-{-    y  =  35. 
2x  -\-    y  =    n. 


9 

-3 

3 

2 

5 

X 

//  +  3 
5 

y 

X-\-4: 
1 

•  %-y 


x-\-y  =  42. 
2y-\-x  =  4A. 
2x-\-y  =  30. 


Elimination  by  Substitution. 

159.  Rule.  Find  the  value  of  one  of  the  unknown 
quantities  in  terms  of  the  other  from  one  equation,  and  sub- 
stitute this  value  in  the  other  equation.  The  latter  will  then 
have  hut  one  u7iknown  quantity. 

Example.  We  take  the  same  example  as  in  the  preced- 
ing method,  namely, 

x-{-    y  =  28, 
3x  -  2?/=  29. 


136  EQUATIONS  OF  THE  FIRST  DEGREE. 

From  the  first  equation  we  have 

xzzz'il^  -  y. 

Substituting  this  value  in  the  second  equation,  it  becomes 

84  -  3.?/  -2y  =  29, 

from  which  we  obtain  as  before 

84-29       ,, 
y  =  —^—  =  ll. 

This  method  may  be  applied  to  any  pair  of  equations  in 
four  ways: 

1.  Find  X  from  the  first  equation  and  substitute  its  value 
in  the  second. 

2.  Find  x  from  the  second  equation  and  substitute  its 
value  in  the  first. 

3.  Find  y  from  the  first  equation  and  substitute  its  value 
in  the  second. 

4.  Find  y  from  the  second  equation  and  substitute  its 
value  in  the  first. 

EXERCISES. 

Solve  the  following  equations  in  four  ways: 


1.  x-}-2y  =  18 

2.  X-    y  =    1 

3.  X  -{-  2y  =  m 


2x  —  y  =  6. 
2x  +  3y  =  14. 
2x  +   y  =  n. 


Elimination  by  Addition  or  Subtraction.  ' 

160.  Rule.  Multiply  each  equation  ly  such  a  factor 
that  the  coefficients  of  one  of  the  unknown  quantities  shall 
hecome  7iumerically  equal  in  the  two  equations. 

Then  hy  adding  or  subtracting  the  equations  we  shall 
have  an  equation  loith  hut  one  unknow?i  quantity. 

Remark.  We  may  take  for  the  factor  of  each  equation 
the  coefficient  of  the  unknown  quantity  to  be  eliminated  in 
the  other  equation,  unless  we  see  that  simpler  multipliers 
will  answer  the  purpose. 

Example.     Taking  again  the  same  equations  as  before, 

x-\-   y  =  2S, 

3x  —  2y  =  29, 

we  multiply  the  first  equation  by  3,  obtaining 

3x  +  Sy  =  84. 


ELIMINATIONS  BY  ADDITION  OR  SUBTRACTION.     187 

The  coefficient  of  x  is  now  the  same  as  in  the  second  given 
equation.     Subtracting  the  second,  we  have 

by  =  55, 
whence  ^  =  11. 

Again,  if  we  multiply  the  first  equation  by  2  and  add  it 
to  the  second,  we  have 

6x  =  85, 
whence  x  =  17. 

Eemark.  We  always  obtain  the  same  result,  whatever 
method  of  elimination  we  use.  But,  as  a  general  rule,  the 
method  of  addition  or  subtraction  is  the  simplest  and  most 
elegant.  Very  often  little  or  no  multiplication  is  necessary. 
Take  the  following  case,  for  instance: 

161.  Peoblem  of  the  Sum  aj;d  Differekce.  The 
sum  and  difference  of  two  numbers  being  given,  to  Ji?id  the 
numbers. 

Let  the  numbers  be  x  and  y. 
Let  «  be  their  sum  and  d  their  difference. 
Then,  by  the  conditions  of  the  problem, 
x^  y  =  s, 
X  —  y  =  d. 
Adding  the  two  equations,  we  have 

2x  =  s  -\-  d. 
Subtracting  the  second  from  the  first, 

2y  =  s  —  d. 
Dividing  these  equations  by  2, 

s  -{-  d      s    ,  d 

_  s  —  d  _  s       d 
y  ~      2      ~  2  ~"  2  * 
We  may  therefore  state  these  conclusions  in  the  form  of 
the  following 

Theorem.  Half  the  sum  of  any  two  numbers  plus  half 
their  difference  is  equal  to  the  greater  number;  and 

Half  the  sum  of  any  tivo  numbers  minus  half  their  differ- 
ence is  equal  to  the  lesser  number. 


138 


EQUATIONS  OF  THE  FIR8T  DEGREE. 


EXERCISES. 

Solve  the  following  pairs  of  equations: 


1.  X  +  2?y  =:  36 

2.  2a;-f    i/ =    8 

3.  dx-  by  =  17 

4.  x-\-2y  =  20 

5.  dx  —  4ty  =    c 

6.  ax  -\-  by  =z  m 
'7.  ax  -\-  by  =  c 
8.  ax  -\-  by  =  p 

X-  6  _y-\-7 
2      ~      3 


9. 
10. 
11. 

12. 


X  —2y  =  24. 
2x  —  y  =  8. 
3x  -i-  6y  =  37. 
2i2;  +  ?/  =r  25, 
2x  -]-'7y=  e. 
ax  —  by  =  n, 
mx  +  ny  =  p. 
mx  —  ny  =   q. 

2a;  -  4      by  - 


ax  -\-by  =  c, 
1; 


a~^  b 


^     I     y 

a  -\-  b      a  —  b 


13.   - 


14.   - 


15. 


x-j-y 

a 

y       b 

2 


771  \ 


y-a 
x-y  _ 


2 
a'x  +  b'y 

•1  _  ^  =  1 
«       ^       2 

a;  —  ^/ii/  +  a 
X  -\-  my 


h. 


y  -. 


=  1, 


3 


16^.  Sometimes  we  may  advantageously  treat  expres- 
sions containing  the  unknown  quantities  as  if  they  were  single 
symbols,  in  accordance  with  Priilciple  II.  of  the  algebraic 
language. 

EXERCISES. 

1.  5{x  +  y)-2{x-y)  =  U;       5(a:  +  ^)  +  2(2:  -  ^z)  =  76. 
Solution.     Taking  the  sum  and  difference  of  the  two  equa- 
tions as  they  stand,  we  have 

10(:?;  +  y)  =z  44  +  76  =  120,  whence  x -{- y  =  12; 
4(0;  —  i/)  =  76  —  44  =    32,  whence  x  —  y  =■  8. 
Finally,  adding  and  subtracting  the  last  pair,  we  have 
2x  =  20,  2y  =  4;  whence  x  =  10,  y  =  2. 

2.  3(x  +  2y)  +  2{x  -  2y)  =  65;    3{x  +  2y)  -  2(x  -  2y)  =  17. 

3.  2(2a;-?/)  +  3(2a:  +  ?/)  =  28;    4(2a;  -  i/)  +  3ferr  +  ?/)  :=  42. 


4.  2(5a:-3i/)  +  (4a;-2/)  =  40;      2 


^x-3jf) 


i^x—y)  --20. 


EqUATIONS  WITH  THREE  UNKNOWN  QUANTITIES.    139 

^'  x~^y       24'        X       y       24* 
Solutioyi.     Adding  and  subtracting  the  equations  as  they 
stand,  we  find 

-  =  —  =  -,  whence  -  —  77  and  a:  =    8; 
a:       24      4'  X        ^ 

=  -,  whence  -  =  kt  and  «/  =  24.  . 


=  3. 


Section  III.     Equations   of  the  First  Degree 
WITH  Three  or  Mo^  Unknown  Quantities. 

163.  When  the  values  W  scA^eral  unknown  quantities  are 
to  be  found,  it  is  necessary  to  have  as  many  equations  as  un- 
known quantities. 

164.  Method  of  Elimmati^.  When  the  number  of  un- 
known quantities  exceeds  two,  the  most  convenient  method 
of  elimination  is  generally  that  by  addition  or  subtraction. 
The  unknown  quantities  are  to  be  eliminated  one  at  a  time 
by  the  following  • 

Rule.  1.  Select  an  unknowit  quantity  to  he  first  elimi- 
nated. It  is  best  to  hegin  ivith  the  quantity  which  appears  in 
the  fewest  equations  or  has  the  simplest  coefficients. 

2.  Select  one  of  the  equations  containing  this  unknown 
auantify  as  an  elfll^altfki  envafu)}^ 


(a) 


140  EQUATIONS  OF  THE  FIRST  DEGREE. 

3.  Eliminate  the  quantity  letween  this  equation  and  each 
of  the  others  in  successi07i. 

We  shall  then  have  a  second  system  of  equations  less  by 
one  in  number  than  the  original  system,  and  containing  u 
number  of  unknown  quantities  one  less. 

4.  Repeat  the  process  on  the  neiu  system  of  equations,  and 
continue  the  repetition  until  only  one  equation  u^ith  one  un- 
known quantity  is  left. 

5.  Having  found  the  value  of  this  last  unknoivn  quantity, 
the  values  of  the  others  can  he  found  hy  successive  substitution 
in  07ie  equation  of  each  system. 

Example.     Solve  the  equations 

(1)  ^x  —  dy  —    z  -\-    u  =  62 

(2)  X  -    y  -^  2z  -^  2u  =  20 

(3)  2x  -{-  2y  -    z-  2u  =    3 

(4)  X  -\-2y  -j-    z  -^    u  =    2. 

We  shall  select  x  as  the  first  quantity  to  be  eliminated,  and 
take  the  last  equation  as  the  eliminating  one.  We  first  mul- 
tiply this  equation  by  three  such  factors  that  the  coefficient 
of  X  shall  become  equal  to  the  coefficient  of  x  in  each  of  the 
other  equations.  These  factors  are  4,  1  and  2.  We  write 
the  products  under  each  of  the  other  equations,  thus: 
(1),  4:X  —  Sy  —    z  -\-    u  =  62, 

(4)  X  4,  4x -]-  Sy  -\-  4:z -{-  4.U  =    8. 

(4)  X  1, 

^4)  X  2, 

By  subtracting  one  of  each  pair  from  the  other,  we  ob- 
tain the  equations 

(1')  lly  -^  6z  -i-Su  =  -  U,) 

(2')  dy  -    z  -    u=  -  18,  y  (b) 

(3')  2y  ^3z-\-4:U  =         1.  ) 

The  unknown  quantity  x  is  here  eliminated,  and  we  have 
three  equations  with  only  three  unknown  quantities.  Next 
we  may  eliminate  y  by  using  the  last  equation  as  the  elimi- 
nating one.     We  proceed  as  follows: 


X  -    y  -\-2z  -\-2u  = 
^  +  ^.y  +    ^  +    ^  = 

20, 
2. 

2x  -i-  2y  —    z  —  2u  = 
2x  +  4i/  -{-  2z  -\-  2u  =: 

3, 
4. 

EQUATIONS  WITH  THREE  UNKNOWN  QUANTITIES.   141 


(1')  X  2, 
(?/)  X  11, 

22^/  +  10^  +    6w  = 
22?/  +  332;  4-  44?^  = 

-  88, 
11. 

(2')  X  2, 
(3')  X  3, 

Qy  -    2z  -    2tc  = 
6y  -\-    dz  -\-  12u  = 

-  36, 
3. 

Subtracting,  we  have 
(1")  232  +  3Su  =  99,  )  .  . 

(2")  11^  +  14w.  =  39,  f  ^"^^ 

a  system  of  two  equations  with  only  two  unknown  quantities. 

From  these  equations  we  find: 

(1")  X  11,  253;^  +  418?^  =  1089, 

(2")  X  23,  253;^  +  322w  =    897. 

96?^  =    192, 
whence  u  =        2. 

Having  thus  obtained  the  value  of  one  unknown  quantity, 
we  find  the  values  of  tlie  remaining  ones  as  follows: 
From  (2")  we  have 

11;^  =  39  -  14?^  =  39  -  28  =  11, 
whence  z  =  1. 

From  (1')  we  have 

Uy  =  -  u  -  6z  -  du 

=  -  44  -  5  -  6  =  -  55, 
whence  ^  =  ~  ^• 

From  (1)  we  have 

4.x  =  62  -\-  Sy  -{-  z  —  u 
•  =  52  -  15  +  1  -  2  =  36, 
whence  a;  =  9. 

Note.  The  student  should  now  verify  these  results  by  substituting 
the  values  of  x,  y,  z  and  u  in  the  four  original  equations  {a)  and  see 
wliether  they  are  all  satisfied. 

EXERCISES. 

1.  One  of  the  best  exercises  for  the  student  will  be  that 
of  resolving  the  previous  equations  (a)  by  taking  the  last 
equation  as  the  eliminating  one,  and  performing  the  elimina- 
tion in  different  orders;  that  is,  begin  by  eliminating  u,  then 
repeat  the  whole  process  beginning  with  z,  etc.  The  final 
results  will  always  be  the  same. 


142  EQUATIONS  OF  THE  FIRST  DEGREE. 

2.  Find  the  values  of  x,  y,  z  and  ti  from  the  equations 

x-\-y-^z-\-u  =  4:a, 
X  -\-  y  —  z  —  u  =  4:b, 
x  —  y  -\-  z  —  u  =  4:C, 
X  —  y  —  z  -\-  u  =  4:cl. 

Remark.  This  exercise  requires  no  multiplication,  but  only  addi- 
tion and  subtraction  of  the  different  equations. 

3.   %x  —  by  -\-^z  =  11,  4.  ax  -\- hy  -\-  cz  =  A, 

dx  -\-2y  -    z  =  31,  a'x  +  %  +  c'z  =  A'\ 

5x-\-3y  -  2z  =  52.  a'x  +  b'y  +  c'z  =  A\ 

Many  of  the  following  equations  can  be  simplified  by  add- 
ing and  subtracting,  so  as  to  reach  a  solution  more  expedi- 
tiously than  by  following  the  general  rule: 


5. 

X  ^  y 

= 

a, 

6. 

X  -\r  y  =  a, 

x^  y  -^  z 

= 

h 

y  ^  z  =  i, 

X  -\-  y  -\-  z  -{-  u 

= 

c, 

z  -\-  X  =  c. 

X  -  y 

— 

d. 

7. 

X  —  7iy  =  a, 
y  —  nz  =  b, 
z  —  nx  ^=  c. 

8. 

y    '   z 

Z          X 

9. 

1  _1^_1 

X       y       c' 
1       i  _  1      - 
y       z  ~  a' 

1  +  1  =  1 

Z  ^  X         h' 

10. 

X  ^  y 

'-  +  ^  =  1. 

Z     '    X 

11. 

X  —  2y          = 

2, 

12. 

ax  -\-  hy  =  c, 

X  -    y  -\-  z  = 

5, 

cy  -\-hz  =  a. 

X  +  2u           = 

7, 

az  -{-  ex  =  h. 

2x  —    z           = 

6. 

13. 

3u  —  X  =  m, 
dx  —  y  =  n, 
Sy  -  z  =p, 
3z  —  y  =  q. 

14. 

mx  —  y  =  Of 
my-  z  =0, 
mz  —  X  •=  a. 

EQUATIONS  WITH  THREE  UNKNOWN  QUANTITIES.   143 

15.  x^y  -^z  =  0, 
(m  +  n)x  +  (71  -{-p)y  -\- (p  +  m)z  =  0, 

7)inx  -j-  npy  -\-  pmz  =  1. 

16.  X  -\-  y  =  2x  —  y  -{-  IS  =  Sx  —  y  —  (5. 

Note.     Equations  of  this  kind,  which  often  trouble  the  beginner, 

art;  easily  solved  by  equating  separately  different  pairs  of  the  equal 

Micnibers,     For  instance,  the  second  and  third  members  are  J 

2x  -  y  -^  IS  =  dx  -  y  -  Q.  i 

By  transposing  the  unknown  quantities  to  the  second  member,  and 

6  to  the  first,  we  have  at  once 

24  =  x    or    X  =  24.  (1) 

The  first  two  members  alone  are 

x-{.y  =  2x  —  y-\-18, 
which  gives  by  transposition 

2y  -  x  =  18, 
from  which  y  is  found  by  (1). 

It  is  to  be  noted  that  in  all  such  cases  tliere  are  as  many  equations  as 
signs  of  equality. 

17.  2x  -dy  -'d  =  dx-2y  -22  =  X-  y^  1. 

18.  xy  —  X  —  y  -\-  24:  =  xy  -j-  x  -{-  y  =  xy  -{-  x  —  3y  -{-  6. 

19.  dx  =  bx  —  y  =:  7x  —  y  —  4:  =  z. 

20.  xy  -}-  X  =  xy  -{- Sx  -^  y  —  28  =  xy  —  3x  +  44. 

21.  2x  =  -  3y  =  5{x  +  ?/'  +  1). 

22.  ax  =  by  =  (a  -  b)  (x  +  y  +  1). 

23.  xy  =  (x  +  5)  (y  -2)=  (x  +  9)  (^  -  3). 

165.  Problems  leading  to  Equations  of  the  First  Degree. 
Ill  the  solution  of  the  following  problems  the  student  will 
sometimes  find  it  convenient  to  use  but  one  unknown  quan- 
tity, and  sometimes  to  use  two  or  more.  He  must  always 
state  as  many  equations  as  there  are  unknown  quantities,  but 
the  method  to  be  followed  in  the  solution  must  be  left  to  his 
own  ingenuity. 

1.  Divide  11.25  between  two  boys,  giving  A  43  cents  more 
than  B. 

Svggestion.     Call  x  A's  share  and  y  B's  share. 

2.  Divide  a  line  85  feet  long  into  two  parts  of  which  one 
shall  be  22  feet  longer  than  the  other. 

3.  Half  the  sum  of  two  numbers  is  95  and  half  their  differ- 
ence is  55.     What  are  the  numbers? 

4.  The  total  vote  for  two  candidates  for   Congress  was 


144  KQUATIONti  OF  THE  FlRiST  DEGREE. 

20,185  and  the  Republican  majority  was  1093.     How  many 
votes  were  cast  for  each  party? 

5.  A  line  being  divided  into  two  parts,  the  whole  line/;//^s 
the  greater  part  is  81  feet,  and  the  whole  line  plus  the  lesser 
part  54  feet.     What  is  the  length  of  the  line? 

6.  Divide  $455  between  two  men,  so  that  ^  the  share  of 
one  shall  be  equal  to  -J  the  share  of  the  other. 

7.  There  were  subscribed  125,000  to  a  college  fund.  A 
subscribed  15000  less  than  B  and  0  together,  and  B  subscribed 
1^2000  more  than  0.     What  did  each  subscribe? 

8.  A  man  has  two  horses,  with  a  saddle  Avorth  $50.  The 
saddle  and  first  horse  are  together  worth  as  much  as  the  sec- 
ond horse;  the  saddle  and  second  horse  are  together  worth 
twice  as  much  as  the  first  horse.     What  is  the  value  of  each? 

9.  A  man  has  a  buggy  and  two  horses  worth  in  all  $800. 
When  the  best  horse  is  harnessed  to  the  buggy  the  team  is 
Avorth  $400  more  than  the  other  horse.  When  the  poorest 
horse  is  harnessed  the  team  is  worth  $100  more  than  the  good 
horse.     What  is  the  value  of  the  bnggy  and  of  each  horse? 

10.  Two  men  have  together  225  acres  of  land  worth 
$15,000.  A's  land  is  worth  $50  per  acre  and  B's  worth  $100 
per  acre.     How  much  land  has  each? 

11.  A  sum  of  $15.60  was  divided  among  72  children,  each 
boy  getting  25  cents  and  each  girl  20  cents.  How  many 
Avere  boys  and  how  many  Avere  girls? 

12.  The  first  of  tAvo  cisterns  has  twice  as  much  Avater  as 
the  second.  If  120  gallons  be  poured  from  the  first  into  the 
second,  the  latter  will  then  have  twice  as  much  as  the  first. 
How  much  has  each? 

13.  What  fraction  is  that  which  becomes  equal  to  ^  Avhen 

its  numerator  is  increased  by  1,  and  to  ^  when  its  numerator 

is  diminished  by  1? 

x 
Suggestion.    Call  —  the  fraction. 

14.  What  fraction  becomes  equal  to  J  both  when  1  is  sub- 
tracted from  its  numerator  and  when  2  is  added  to  its  de- 
nominator? 

15.  A  and  B  each  had  a  certain  sum  of  money.  A  got 
$25  more  and  thus  had  twice  as  much  as  B.     Then  B  got 


PI10BLEM8.  145 

^IO'k)  more  and  had  twice  as  much  as  A.      How  much  had 
each  at  first  ? 

16.  In  a  Congressional  election  17,346  votes  were  cast^ 
the  Democrat  getting  2432  more  than  the  National,  and  the 
RepubUcan  878  more  than  the  Democrat.  What  Avas  the 
vote  of  each? 

17.  A  man  is  7  years  older  than  his  wife,  and  10  years 
hence  his  age  will  be  double  what  his  wife's  was  10  years  ago. 
What  are  their  present  ages? 

Suggestion.  If  we  call  x  the  wife's  present  age,  what  is  the  man's 
present  age?  What  will  be  his  age  10  years  hence?  What  were  tlieir 
respective  ages  10  years  ago? 

18.  A  boy  is  now  half  the  age  of  his  elder  brother,  but  in 
24  years  he  will  be  f  his  age.     What  are  their  present  ages? 

19.  The  combined  ages  of  a  man  and  his  wife  are  now  62 
years,  and  in  11  years  he  will  be  older  than  she  will  by  ^  of 
her  age.     What  are  their  present  ages? 

20.  Two  men  having  engaged  in  gambling,  A  won  14  from 
B  and  then  had  twice  as  much  money  as  B.  Then  B  won 
$25  and  their  shares  were  equal.    How  much  had  each  at  first  ? 

21.  A  sum  of  money  being  equally  divided  between  A  and 
B,  A  got  $75  more  than  ^  of  it.     What  was  the  sum  divided? 

22.  What  is  the  length  of  that  line  which  being  divided 
into  three  equal  parts,  each  part  is  2  inches  longer  than  one 
fourth  of  the  line? 

23.  A  sum  of  $520  being  divided  between  A,  B  and  C,. 
B's  share  was  f  of  A's  share  and  $40  more  than  C's  share. 
What  was  the  share  of  each? 

24.  Another  sum  being  divided,  A  got  $40  less  than  halL 
B  $40  more  than  one  fourth,  and  0  $40  more  than  one  sixth. 
What  was  the  amount  and  the  share  of  each? 

25.  Two  men  had  between  them  96  dollars.  A  paid  B 
one  fourth  of  his  (A's)  money,  but  B  lost  $6  and  then  had 
twice  as  much  as  A.     How  much  had  each  at  first? 

26.  Two  cisterns  being  each  partly  full  of  water,  ^  of  the 
water  in  the  first  was  poured  into  the  second,  which  then  had 
120  gallons.  Then  \  of  what  was  left  was  poured  from  the 
first  into  the  second,  when  the  latter  had  twice  as  much  as  the 
first.     How  much  did  each  contain  at  first? 


146  EQUATIONS  OF  THE  FIRST  DEGREE. 

27.  At  a  city  election  Jones  had  a  majority  of  244  votes 
over  Smith.  But  it  was  found  that  ^  of  Jones's  votes  and 
yig-  of  Smith's  votes  were  illegal,  and  on  correcting  this 
Smith  had  a  majority  of  77.  What  was  the  legal  vote  of 
each?  Ans.  Smith,  10,458;  Jones,  10,381. 

Suggestion.  Take  the  whole  numbers  of  votes  first  cast  as  the  un- 
known quantities. 

28.  An  apple- woman  bought  a  lot  of  apples  at  5  for  2  cents. 
She  sold  half  at  2  for  a  cent,  and  half  at  3  for  a  cent,  thus  gain- 
ing 10  cents.     How  many  apples  were  there?        Ans.  600. 

29.  A  train  ran  half  the  distance  between  two  cities  at 
the  rate  of  30  miles  an  hour  and  half  the  distance  at  50  miles 
an  hour.  It  performed  the  return  journey  at  the  uniform 
speed  of  40  miles  an  hour,  thus  gaining  half  an  hour  on  its 
time  in  going.     What  was  the  distance?       Ans.  300  miles. 

Remark.  In  all  questions  involving  time,  constant  mlocity  and  dis- 
tance we  have  the  fundamental  relation:  Distance  =  'oelocity  X  time. 

30.  A  man  bought  29  oranges  for  a  dollar,  giving  3  cents 
a  piece  for  poor  ones  and  4  cents  for  good  ones.  How  many 
of  each  kind  had  he? 

31.  A  grocer  bought  a  lot  of  sugar  at  8  cents  a  pound  and 
of  coffee  at  12  cents  a  pound,  paying  $8.60  for  the  whole. 
He  sold  the  sugar  at  10  cents  a  pound  and  the  coffee  at  14 
'Cents,  realizing  $10.40.     How  much  of  each  did  he  buy? 

32.  A  huckster  bought  a  lot  of  oranges  at  1  cent  each  and 
lemons  at  2  cents  each,  paying  $1.45  for  the  lot.  5  of  the 
oranges  and  10  of  the  lemons  were  bad,  but  he  sold  the  good 
fruit  at  2  cents  each  for  oranges  and  3  cents  for  lemons,  real- 
izing $1.90.     How  many  of  each  kind  did  he  buy? 

33.  The  sum  of  the  ages  of  two  brothers  is  now  4  times 
tlie  difference,  but  in  6  years  it  will  be  5  times  the  difference. 
What  are  their  present  ages? 

34.  For  $6  I  can  buy  either  4  pounds  of  tea  and  20  pounds 
of  coffee  or  2  pounds  of  tea  and  25  of  coffee.  What  is  the 
price  per  pound  of  each? 

35.  An  almoner  had  3  equal  sums  of  money  to  divide  be- 
tween 3  families.  The  second  family  had  5  more  than  the 
first  and  so  got  $2  apiece  less,  and  the  third  had  4  more 
than  the  second  and  got  $1  a|)iece  less.  What  were  the 
numbers  of  the  families  .-md  tlic  equal  ;nTir)unts  divirlorl? 


riWBLEMS.  147 

36.  If  A  can  do  a  piece  of  work  in  3  drys  and  B  in  *: 
days,  in  what  time  can  they  do  it  if  both  work  together? 

Suggestion.  In  one  day  A  can  do  \  and  B  can  do  |.  Hence  both 
together  can  do  ^  +  i-     But  if  x  be  the  time  in  which  both  can  do  it, 

they  can  do  —  of  it  in  a  day.     Hence 


-  =  -  +  - 

37.  If  one  pipe  can  lill  a  cistern  in  12  minntcs  and  ar.other 
in  18  minutes,  in  what  time  can  they  both  fill  it? 

38.  A  cistern  can  be  emptied  by  two  faucets  in  12  minutes 
and  by  one  of  them  in  36  minutes.  In  what  time  can  it  be 
emptied  by  the  other? 

39.  A  cistern  can  be  filled  by  a  pipe  in  25  minutes  when 
the  faucet  is  left  running,  and  in  15  minutes  when  the  faucet 
is  closed.     In  what  time  will  the  faucet  empty  it? 

40.  Three  men  can  together  perform  a  piece  of  work  in  12 
days.  A  can  do  twice  as  much  as  B,  and  B  twice  as  much 
as  0.     In  what  time  could  each  one  separately  do  the  work? 

41.  A  and  B  can  together  do  a  job  of  work  in  8  days,  B 
and  0  in  9  days,  and  0  and  A  in  12  days.  In  what  time  can 
each  one  alone  do  it?     In  what  time  can  they  all  do  it? 

Ans.  A,  20f ;  B,  Id^  ;  0,  28f ;  all,  6^V 

42.  A  train  performs  the  journey  from  Washington  to 
Chicago  in  a  certain  time  at  a  certain  speed.  By  going  16 
miles  an  hour  faster  it  gains  9  hours,  and  by  going  8  miles 
an  hour  slower  it  takes  9  hours  longer.  What  is  the  original 
time  and  speed  and  what  the  distance? 

43.  A  privateer  sights  an  enemy's  ship  9  miles  away,  flee- 
ing at  the  speed  of  6  miles  an  hour.  If  the  privateer  chases 
her  at  the  rate  of  8  miles  an  hour,  in  what  time  and  at  what 
distance  will  she  overtake  her? 

Method  of  Solution.  Let  t  be  the  time  and  d  the  distance.  Because 
the  privateer  sails  8  miles  an  hour,  we  have  d  =  8t.  The  ship  having 
9  miles  less  to  go  will,  when  overtaken,  have  sailed  {d  —  9)  miles. 
Therefore  d  -  9  =  6t.     From  these  two  equations  we  find  d  and  t. 

44.  A  man  gets  into  a  stage-coach  driving  7  miles  an 
hour  for  a  pleasure-ride;  but  he  must  walk  home  at  the  rate 
of  3  miles  an  hour,  and  be  gone  only  5  hours.  How  far  can 
he  ride? 


148  EQUATIONS   OF  THE  FIliST  DEGREE. 

45.  A  train  performed  a  journey  in  6  hours,  going  one 
third  the  way  at  the  rate  of  30  miles  an  hour  and  two  thirds 
at  the  rate  oi'  40  miles  an  hour.     What  was  the  distance? 

46.  A  man  had  two  casks  of  wine  containing  together  75 
gallons.  He  poured  one  fourth  the  contents  of  the  first  into 
the  second,  and  then  poured  one  third  the  increased  contents 
of  the  second  cask  into  the  first.  The  first  then  contained  'Z^ 
gallons  more  than  the  second.  How  much  did  each  contain 
at  first? 

47.  Two  casks  contain  between  them  s  gallons  of  wine, 
and  one  had  d  gallons  more  than  the  other.  How  much  wine 
had  each? 

48.  A  board  is  divided  into  two  parts.  The  whole  board 
plus  the  greater  part  is  m  feet  and  the  Avhole  board  plus  the 
lesser  part  is  n  feet.     What  is  the  length  of  the  board? 

49.  Divide  a  board  I  feet  long  so  that  two  thirds  of  one 
part  shall  be  equal  to  one  half  the  other  part. 

50.  A  grocer  mixed  k  pounds  of  tea  worth  j(?  cents  a  pound 
with  Ji  pounds  worth  n  cents  a  pound.  How  much  per  pound 
was  the  mixture  w^orth? 

Note.  The  solution  of  this  question  does  not  really  require  any 
equation. 

51.  A  grocer  has  t  pounds  of  tea  worth  r  cents  a  pound, 
which  is  formed  by  mixing  two  kinds,  one  worth  p  cents  a 
pound  and  the  other  q  cents  a  pound.  How  much  of  each 
kind  did  he  mix? 

52.  A  boy's  age  is  now  one  third  that  of  his  elder  brother, 
but  in  t  years  it  will  be  one  half.  What  are  their  present 
ages? 

53.  Three  casks  contain  altogether  on  gallons  of  wine. 
By  pouring  a  gallons  from  the  first  into  the  second  and  then 
h  gallons  from  the  second  into  the  third,  the  quantities  in  the 
three  casks  become  equal.  How  much  did  each  cask  con- 
tain at  first? 

54.  A  man  who  must  be  back  in  t  hours  starts  in  a  coach 
going  m  miles  an  hour  and  walks  back  at  the  rate  of  7i  miles 
an  hour.     How  far  can  he  go  and  get  back  in  time? 

55.  If  one  man  can  do  a  piece  of  work  in  a  days  and 
another  in  l  days,  in  what  time  can  they  both  do  it? 


INTERPRETATION  OF  NEGATIVE  RESULTS.       149 

56.  If  A  can  do  the  work  in  a  days,  B  in  0  days  and  C  in 
c  days,  in  what  time  can  they  all  do  it  if  work  together? 

57.  If  A  and  B  together  can  do  a  piece  of  work  in  7n  days 
and  A  alone  in  a  days,  in  what  time  can  B  alone  do  it? 

58.  A  man  bonglit  tea  and  coffee,  p  pounds  in  all,  for  r 
cents,  giving  m  cents  a  pound  for  tea  and  n  cents  a  pound  for 
coffee.     How  much  of  each  did  he  buy? 

To  prove  the  results  add  the  two  amounts  and  see  wlicther  tliey 
make  p  pounds. 

59.  Divide  m  dollars  among  3  men,  giving  B  a  dollars 
more  than  0,  and  A  b  dollars  more  than  B. 

GO.  If  a  train  makes  half  its  journey  at  the  rate  of  vi 
miles  an  hour  and  the  other  half  at  the  rate  of  n  miles  an 
hour,  what  is  its  average  speed? 


Interpretation  of  Negative  Results. 

166.  An  answer  to  a  problem  may  sometimes  come  out 
negative.  This  shows  that  the  answer  must  be  reckoned  in 
the  opposite  direction  from  that  assumed  as  positive  in  the 
enunciation  of  the  problem. 

Example.  A  man  is  30  years  old  and  his  wife  is  24.  In 
how  many  years  will  he  be  half  as  old  again  as  she  is? 

Solution.  Let  us  put  t  for  the  required  number  of  years. 
In  t  years  his  age  will  be  30  +  if  and  hers  will  be  24  +  t. 

Because  the  conditions  of  the  problem  require  his  age  to 
be  half  as  much  again  as  hers,  we  have 

30  +  ?f  =  |(24  +  0- 

Solving  this  equation,  we  find  ^  =  —  12. 

This  negative  result  shows  that  the  time  when  the  re- 
quired condition  was  fulfilled  is  not  m  the  future  but  in  the 
past,  when  the  wife  was  12  and  the  man  18. 

Had  we  stated  the  problem.  How  many  years  ago  was  his 
age  half  as  much  again  as  hers?  t  would  have  had  the  opposite 
sign  all  the  way  through  and  would  have  come  out  +12 
years,  because  in  the  enunciation  years  past  would  then  be 
regarded  as  positive.     Therefore  whether  time  in  the  future 


J5()  F.qUATJO^'S  OF  THE  FIRST  DEOBEE. 

shall  be  positive  or  negative  depends  on  what  we  assume  as 
positive  in  the  enunciation  of  the  problem. 

Ex.  2.  A  man  is  4  years  older  than  his  wife,  and  6  times 
his  age  is  equal  to  7  times  hers.  How  many  years  ago  was 
7  times  his  age  equal  to  8  times  hers? 

Soluho7i.  Let  us  j^ut  x  for  lier  present  age  and  t  for  the 
required  number  of  years  ago.  Then  his  age  will  be  a;  -j-  4 
vears,  and  by  the  conditions  of  the  problem 

iS{x  +  4)  r=  7x, 
But  t  years  ago  his  age  must  have  been  x  -{-  4:  —  t.  and  hers 
must  have  been  x  —  t.     By  the  conditions  of  the  problem 

7(^  +  4  -  t)  =  8{x  -  t). 
Reducing  and  solving  these  two  equations,  we  obtain 

x=       24, 
t=  ~    L 
Here  t  comes  out  negative  as  before;  but  it  does  not  mean 
years  ago  as  it  did  in  the  first  example,  but  years  in  the  future, 
because  in  stating  the  problem  we  assumed  years  in  the  past 
to  be  positive. 

16T.  Problem,  of  the  Couriers.  A  courier  starts  from  his 
station  riding  8  miles  an  hour.  4  hours  afterwards  he  is  fol- 
lowed by  another  riding  10  miles  an  hour.  How  long  will  it 
require  for  the  second  to  overtake  the  first,  and  wliat  will  be 
the  distance  travelled? 

If  t  be  the  number  of  hours  required,  the  first  will  have 
travelled  t  -\-  ^  hours  and  the  second  t  hours  when  they  come 
together.  Because  one  goes  8  miles  an  hour  and  the  other 
10,  the  whole  distance  travelled  by  the  one  will  be  8(^  -j-  4), 
and  that  travelled  by  the  other  will  be  10^.  Because  these 
distances  are  equal,  we  have 

8if  +  32  =  l^t 
Solving  this  equation,  we  have  t  =  16,  whence  distance  =  160. 

Now  let  us  change  the  problem  thus:  The  first  courier 
still  riding  8  miles  an  hour,  the  second  starts  out  after  him 
in  2  hours  and  travels  6  miles  an  hour.  In  what  time  and  at 
what  distance  will  they  be  together? 

Treating  the  equation  in  the  same  way  as  before,  we  find 
the  equation  to  be 

8(t  +  2)  =  6/ 


INTEUPHKTATION  OF  NEGATIVE  BESifLTti.        l|)i 

Solving  this  equation,  the  result  is 
t  =  -    S, 
d=  -  48. 

These  answers  being  negative  show  that  there  is  no  point 
on  the  road  in  the  positive  direction,  and  no  time  in  the 
future  when  the  couriers  would  be  together. 

Thus  far  the  algebraic  result  agrees  with  common-sense; 
but  common-sense  seems  to  say  that  the  couriers  would  never 
be  together,  whereas  the  algebraic  answer  gives  a  negative 
time  and  a  negative  distance  on  the  road  when  they  were 
together. 

The  explanation  of  the  difficulty  is  this.  Supposp,  8  to  be 
the  point  from  which  the  couriers  started,  and  AB  ^he  road 
along  which  they  travelled  from 

S  toward  B.     Suppose  also  that  i  — ^^ 

the  first  courier  started  out  from 

S  at  8  o'clock  and  the  second  at  10  o'clock.  By  the  rule 
of  positive  and  negative  quantities,  distances  towaids  A  are 
negative.  Now,  because  algebraic  quantities  do  not  com- 
mence at  0,  but  extend  in  both  the  negative  and  positive 
directions,  the  algebraic  problem  does  not  suppose  the  cour- 
iers to  have  really  commenced  their  journey  at  S,  but  to  have 
come  from  the  direction  of  A,  so  that  the  first  one  passes  ^S', 
without  stopping,  at  8  o'clock,  and  the  second  at  10.  It  is 
plain  that  if  the  first  courier  is  travelling  the  faster,  he  must 
have  passed  the  other  before  reaching  S\  that  is,  the  time 
and  distance  are  both  negative,  just  as  the  problem  gives 
them. 

The  general  principle  here  involved  may  be  expressed  thus: 

In  algebra,  roads  and  journeys,  like  time,  have  no  hegi7i- 
ning  and  no  end. 

168,  In  the  following  problems,  when  negative  results 
are  obtained,  the  pupil  should  show  hoAv  they  are  to  be  in- 
terpreted. 

1.  Albany  is  6  miles  below  Troy  on  the  Hudson  River. 
Let  us  suppose  the  river  to  flow  at  the  rate  of  4  miles  an 
hour.  How  fast  must  a  man  starting  from  Troy  roAV  his 
boat  through  the  water  in  order  that  he  may  reach  Albany  in 
1  hour? 


152  EQUATIONS  OF  THE  FIRST  DEGREE. 

2.  The  same  thing  being  supposed,  how  fast  must  he  row 
in  order  to  reach  Albany  in  2  hours?  Interpret  the  negative 
result. 

3.  If,  starting  from  Albany,  he  rows  up  the  river  at  the 
rate  of  6  miles  an  hour,  how  long  will  it  take  him  to  reach 
Troy? 

4.  If  he  rows  at  the  rate  of  3  miles  an  hour,  how  long 
will  it  take  him  to  reach  Troy?     Explain  the  negative  result. 

Apply  the  principle  laid  down  by  tlie  diagram  in  tiie  preceding 
section. 


Memoeanda  for  Review. 

Define :  Equation  of  the  first  degree;  System  of  simul- 
taneous equations;  Elimination. 

Equations  of  the  First  Degree  luith  One  Unhioivn  Quantity. 
Give  rule  for  solution. 


Two   Unknoivn  Quantities, 

[Comparison;  Rule. 
\  Substitution;  Rule. 


Ellmma-    (  Comparison;  Rule. 

^     (  Addition  and  subtraction;  Rule. 

Theorem  of  the  Sum  and  Difference. 
Explain  theorem. 

Three  or  More  Unknown  Quantities. 
Method  of  elimination;  Rule, 


CHAPTER  IV. 

RATIO    AN  D    PROPORTION 


Section  I.     Of  Ratio. 


169.  Def.  When  a  greater  quantity  contains  a 
lesser  one  an  exact  number  of  times,  the  greater  is 
said  to  be  a  multiple  of  the  lesser,  and  the  lesser  is 
said  to  measure  the  greater. 

Example.     If  the  line  a  ^      ' ' 

contains   the   line   b   exactly 

three  times,  «  is  a  multiple  6 

of  l,  and  b  measures  a. 

170.  Def.  When  two  quantities  may  each  be 
measured  by  the  same  lesser  quantity,  they  are  said 
to  be  commensurable,  and  the  lesser  quantity  is  said 
to  be  their  common  measure. 

Example.     If  the  line  A  ,    ^    ,    5    i    ^>    i    &    i    b    i 

contains   the    measure   b   five 
times,  and  B  contains  it  three  \    h    \    h    \    h   \ 

times,  A  and  B  are  cornmen-        ^  " 

.surable,  and  b  is  their  common  measure. 

171.  Def.  The  ratio  of  two  niafrnitudes  is  the 
number  by  which  one  must  be  multiplied  to  j)roduce 
the  other. 

Example.  In  the  first  fio^ure  ahove  the  ratio  of  a  to  b  is 
3,  because  b  X  S  =  a. 

173.  Belation  of  Batio  to  Quotient.  To  divide  a  ((imn- 
tity  by  a  number  means  to  separate  it  into  that  number  of 
erpial  parts.  The  divisor  is  then  a  number,  and  the  (|U()tient 
is  a  quantity  of  the  same  kind  as  the  dividend.  For  instance, 
if  we  divide  a  line  by  3,  the  result  will  be  a  line  one  third  as 
long;  if  we  divide  weight  by  4,  the  result  will  be  weight  one 
fourth  as  heavy. 


154  RATIO  AND  PROPORTION. 

But  when  we  measure  one  line  by  another  or  one  weight 
by  another,  the  result  is  neither  a  line  nor  a  weight,  but  a 
number;  namely,  the  number  by  which  one  must  be  multi- 
plied to  produce  the  other.  This  number  is  the  ratio.  Hence 
another  definition: 

173.  A  ratio  is  the  quotient  of  two  quantities  of 
the  same  kind. 

A  ratio  is  expressed  by  writing  the  divisor  after  the  divi- 
dend with  the  sign  :  between.     Thus 

means  the  ratio  of  «  to  ^  =:  3,  or  Z>  X  3  produces  a,  or  a 
divided  by  J  =  3. 

Hence  from  the  equation 

we  have  both  a  :  h  =  ^    and    ^  =  b. 

174.  Def.  The  antecedent  of  the  ratio  is  the 
quantity  divided. 

The  consequent  is  the  divisor. 

Example.  In  the  ratio  a  :  b,  a  is  the  antecedent  and  b 
the  consequent. 

Def.  Antecedent  and  consequent  are  called  terms 
of  the  ratio. 

175.  When  the  consequent  measures  the  ante- 
cedent, the  ratio  is  an  integer. 

When  the  consequent  and  antecedent  are  com- 
mensurable, the  ratio  is  a  vulgar  fraction. 

Example.  In  the  example  of  §  lo9,  because  the  ante- 
cedent contains  the  consequent  3  times  the  ratio  is  the  in- 
teger 3. 

But  in  ti)e  exjiniple  of  g  170,  to  find  the  ratio  A  :  B,  wo 
hn^'o  to  divide    U  into  )>  parts  and  take  5  of  these  parts  to 

make  A,     Hence  the  ratio  of  ^  to  7^  is  '—,  or 

o 


RATIO. 


155 


The  fractional  ratio  —  has  for  its  numerator  the  number 

o 

of  times  the  common  measure  is  contained  in  the  antecedent, 
and  for  its  denominator  the  number  of  times  the  common 
measure  is  contained  in  the  consequent.  Hence  if  we  divide 
botli  terms  into  equal  parts,  we  have  the  theorenj: 

A  ratio  is  equal  to  the  quotient  of  the  number  of  parts 
i?i  the  a7itecedent  divided  hy  the  nmnber  of  equal  parts  i7i  tlie 
consequent. 

176.  When  the  two  terms  of  the  ratio  have  no 
common  measure  they  are  said  to  be  incommensur- 
able. 

An  incommensurable  ratio  cannot  be  exactly  expressed  as 
a  vulgar  fraction,  but  we  can  always  find  a  vulgar  fraction 
which  shall  be  as  near  as  we  please  to  the  true  value  of  the 
ratio,  though  never  exactly  equal  to  it. 

Reaso7i.  Let  us  divide  the  consequent  into  any  number  n 
of  equal  parts.  Measuring  the  antecedent  with  one  of  these 
l)arts  we  shall,  when  the  ratio  is  incommensurable,  always 
have  a  piece  of  the  part  left  over.  By  dropping  this  piece 
from  the  antecedent  it  Avill  become  commensurable.  Xow  by 
making  the  number  7i  of  parts  great  enough,  we  may  make 
each  part,  and  therefore  the  piece  left  over,  as  small  as  we 
l)lease.     For  example: 

When  n  >  100,  piece  over  <       ^\^     of  consequent; 

When  n  >        1000,  piece, over  <     y-J^^     of  consequent: 

When  n  >  1000000,  piece  over  <  xo-o^o-^  ^^  consequent; 
etc.  etc. 


EXERCISES. 

Here  are  four  Hues  

of    different    leugths, 

a,  h,  c  and   d.     Find  

as  nearly  as  you  can,  

by  dividing  up  the  lines,  the  following  ratios: 

1.  a  :h.     Ans.  2.  a  :  c.     Ans. 

3.  a  :  d.     Ans.  4.  b  :  c.     Ans. 

5.  b  :  d.     Ans.  6.   c  \  d.     Ans. 


156  RATIO  AND  PROPORTION. 

Proi)erties  of  Ratios. 

177.  Def.  If  we  intercliange  the  terms  of  a  ratio, 
the  result  is  called  the  inverse  ratio. 

That  is^  ^  :  ^  is  the  inverse  of  A  :  B. 

If  B:A  =  -, 

n 

then  B  =  — A, 

n 

fix 
and  we  have,  by  dividing  by  — , 

fi 

7)1 

or  A  :  B  =  —. 

m 

Because  —  is  the  reciprocal  of  — ,  we  conclude: 
m  n 

Theorem  I.  The  inverse  ratio  is  the  reciprocal  of  the 
direct  ratio. 

178.  Theorem  IT.  If  hoth  terms  of  a  ratio  he  midti- 
2)lie(l  hy  the  same  factor  or  iliridcd  Inj  the  same  divisor,  tite 
ratio  is  not  altered. 

Proof.         Ratio  of  B  to  A  =  B  :  A  =  --. 

A 

If  m  be  the  factor,  then 

Ratio  of  7nB  to  mA  =  mB  :  mA  =  — j  =  -;-, 

mA       A 

the  same  as  the  ratio  of  ^  to  ^. 

Again,  if  both  terms  are  divided  by  tlie  same  divisor, 
this  operation  amounts  to  dividing  both  terms  of  the  frac- 
tion which  expresses  tlie  ratio,  and  so  leaves  the  value  of 
the  fraction  unaltered. 

179.  Theorem  III.  If  both  terms  of  a  ratio  be  increased 
by  the  same  quantity,  the  ratio  will  be  increased  if  it  is  Jess 
than  1,  ajid  diminished  if  it  is  greater  than  1;  that  is,  it  will 
be  brought  nearer  to  unity. 


BATIO.  157 

Example.  Let  the  original  ratio  be  2  :  5  =  |.  If  we  repeatedly  add 
1  to  both  numerator  and  denominator  of  the  fraction,  we  shall  have  the 
series  of  fractions 

2       3       4       5       pfp 
5'     6»     7»     8>     ^^^'i 

each  of  which  is  greater  than  the  preceding,  because 

I  -  I  =  a'a;       whence      |  >  |. 

I  -  I  =  A;        whence      |  >  |. 

8  —  V  ■--  5^;        whence      f  >  f, 

etc.  etc. 

General  Proof.  Let  a  :  b  ha  the  original  ratio,  and  let 
both  terms  be  increased  by  the  quantity  u,  making  the  new 
ratio  a-\-2i  :  h  -\-  n.    The  new  ratio  mimis  the  old  one  will  be 

a-\-  u      ci  _{b  —  '^)?^ 

h  -[-u  ~  b"  b'  +  bu  ' 
If  b  is  greater  than  a,  this  quantity  wll  be  positive,  show- 
ing that  the  ratio  is  increased  ])y  adding  u.    If  b  is  less  than  a, 
tlie  quantity  will  be  negative,  showing  that  the  ratio  is  dimin- 
isbed  by  adding  u. 

EXERCISES. 

1.  What  is  the  ratio 

(a)  of  27  inches  to  3  feet?     Ans.  f. 

(b)  of  1122  feet  to  half  a  mile? 

(c)  of  3  miles  to  2244  yards? 

2.  What  is  the  ratio 

(a)  of  I  of  an  inch  to  -f^  of  a  foot? 
(/;)  of  I  of  an  inch  to  f  of  an  inch? 
(r)   of  I  of  a  yard  to  f  of  a  foot? 

3.  Which  ratio  is  the  greater, 

(a)  3  :  5  or  5  :  8? 

(b)  5  :  3  or  8  :  5? 

(c)  G  :  7  or  7  :  8? 

(d)  7  :  4  or  14:8? 

4.  The  ratio  of  two  numbers  is  f,  and  if  each  of  them  is 
increased  by  4  their  ratio  wdll  be  f .     Find  the  numbers. 

Note.     If  we  call  x  and  y  the  numbers,  the  fact  that  their  ratio  is 

I  is  expressed  by  tlie  equation 

X  _d 

2/  ~  5' 

x-\-  4 
When  each  number  is  increased  by  4  tlie  ratio  will  be  equal  to  — -  .. 

^  ^  yi-4 


158  RAllO  AND  PROPORTION. 

5.  If  1   metre  =  39.37  inches,  what  is   the  ratio  of  10 
metres  to  1270  feet? 

6.  The  ratio  of  two  numbers  is  — ,  and  if  eacli  of  them 

71 

be  increased  by  a  their  ratio  will  become  — .  Find  the  num])ers. 

.  .      .     ^ 

7.  If  the  ratio  of  two  quantities  is  |,  what  is  the  ratio  of 

tAvice  the  antecedent  to  three  times  the  consequent?  What 
is  the  ratio  of  m  times  the  antecedent  to  n  times  the  conse- 
quent? 


Sectiois-  II.     Proportion^. 

180.  Bef.  Proportion  is  an  equality  of  two  or 
more  ratios  whose  terms  are  written. 

Since  each  ratio  has  two  terms,  a  proportion  must  liave  at 
least  four  terms. 

Bef.  The  terms  which  enter  into  the  two  equal 
ratios  are  called  terms  of  the  proportion.. 

If  «  :  Z>  be  one  of  the  ratios  and  p  :  q  the  other,  the  pro- 
portion will  be 

a:h=p'.q.  (1) 

A  proportion  is  sometimes  written 

a  :h  ::  p  :  q, 
which  is  read,  "  As  a^  is  to  h,  so  is  p  to  9."     The  first  form  is  to  be  pre- 
ferred, because  no  other  sign  than  that  of  equality  is  necessar}^  betM^een 
the  ratios;  but  the  equation  may  be  read,  "  As  ^  is  to  b,  so  is  p  to  q," 
whenever  that  expression  is  the  clearer. 

Def.  The  first  and  fourth  terms  of  a  proportion 
are  called  the  extremes;  the  second  and  third  are 
called  the  means. 

Theorems  of  Proportion. 

181.  Theorem  I.  In  a  proportion  the  product  of  the 
extremes  is  equal  to  the  product  of  the  means. 

Proof.  Let  us  write  the  ratios  in  the  proportion  (1)  in 
the  form  of  fractions.     It  will  give  the  equation 

:-  =  f  ^'} 


PBOPORTION,  159 

Multiplying  both  members  of  this  equation  by  Iq,  we  shall 

have 

aq  =  hp.  (3) 

183.  Def.  A  proportion  is  said  to  be  inverted 
when  antecedents  and  consequents  are  interchanged. 

Theorem  IL    A  proportion  remains  true  after  incersion. 
Proof,     If  a  \l)  —  p  \  q, 

,,  a       p 

then  —  =z  =!-' 

0        q 

and  by  taking  the  reciprocal  of  each  member, 

1  =  1 

a       p^ 
whence  h  :  a  —  q  :  p. 

183.  Theorem  III.  If  the  means  in  a  proportion  he  in- 
terchanged, the  proportion  will  still  he  trite. 

Proof  Divide  the  equation  (3)  by  p^q.  We  shall  then 
have,  instead  of, the  proportion  (1), 

a  _h 
p~  q' 
or  a  :  p  =  h  '.  q. 

Def.  The  proportion  in  which  the  means  are  in- 
terchanged is  called  the  alternate  of  the  original  pro- 
l^ortion. 

The  following  examples  of  alternate  proportions  should  be  studied, 
and  the  proof  of  the  equations  proved  by  calculation: 

1:2=    4:8;       alternate,     1:4    =2:8. 
2:3=6:9;  ''  2:6=3:9. 

5  :  2  =  25  :  10;  "  5  :  25  =  2  :  10. 

184.  Theorem  IV.  If  in  a  proportion  we  increase  or 
diminish  each  antecedent  hy  its  consequent,  or  each  consequent 
by  its  own  antecedent,  the  proportion  will  still  he  true. 

Example.     In  the  proportion 

5  :  2  =  25  :  10 
the  antecedents  are  5  and  25,  the  consequents  2  and  10  (§  174).     Increas- 
ing each  antecedent  by  its  own  consequent,  \\\e  proportion  will  be 
5  +  2  :  2  =  25  -4-  10  :  10,        or        7  :  2  =  35  :  10. 


160  BATIO  AND  PROPORTION. 

Diminishing  each  antecedent  by  its  consequent,  tlie  proportion  will 
become 

5  -  2  :  2  =  35  -  10  :  10,        or        3  :  2  =  15  :  10. 
Increasing  each  consequent  by  its  antecedent,  the  proportion  will  be 

5  :  2  +  5  =  25  :  10  +  25,         or        5  :  7  :=  25  :  35. 
These  equations  are  all  to  be  proved  numerically. 

General  Proof.     Let  us  put  the  proportion  in  the  form 

If  we  add  1  to  each  member  of  this  equation  and  reduce, 
it  will  become 

a  +  h  _  p^q^ 

b      ~~T"'  ^^ 

that  is,  a  -\-  J)  \  b  =  p  -\-  q  '.  q.  (6) 

In  the  same  way,  by  subtracting  1  from  each  side,  we  have 

h      -       q     '  ^^^ 

or  a  —  h  \  h  =  p  —  q  \  q.  (8) 

If  we  inv^ert  the  fractions  in  equation  (4),  the  Latter  will 
become 

a       p ' 
By  adding  or  subtracting  1  from  each  member  of  this 
equation,  and  then  again  inverting  the  terms  of  the  reduced 
fractions,  we  shall  find 

a  -.h  ^  a  =  p  :q  ^p\  (9) 

a  :  h  —  a  =^  p  '.  q  —  p.  (10) 

The  forms  (7)   and   (9),  in  which  each  pair  of  terms  is 

added,  are  said  to  be  formed  by  composition.     The  forms  (8) 

and  (10)  are  said  to  be  formed  by  division. 

185.   Composition  and  Division.     Taking  the  quotient 
of  (5)  by  (7),  we  have 

a  —  h      p  —  q^ 
that  is,  a  -\-  J)  :  a  —  1)  =  p  -\-  q  \  p  —  q. 

Def.     This  form  is  said  to  be  formed  from  a  ;  h  =  p  :  q 
by  composition  and  division. 


PROPORTION. 


161 


EXERCISES. 

1.  From  the  proportion 

2  :7 

=  6  :21 

form  as  many  more  proport 

ions  as  you 

can  by  alternation, 

inversion,  composition  and  d 

i  vision: 

Ans.  Alt. 

2:6=7 

:  21. 

Inv. 

7:2  =  21 

:    6; 

a 

6  :2  =  21 

:    7. 

Comp. 

9  :  7  =  27 

:21; 

a 

2:9=6 

:27. 

Div. 

5  :  7  =  15 

:21; 

a 

2:5=    6 

:  15. 

Comp.  and  div. 

9  :  5  =  27 

:  15. 

2.  Do  the  same  with  tlie  proportion 
24  :  9  =  8  :  3. 

186.  1'heorem  \.     If  each   term   of  a  proportion   he 

raised  to  the  same  poiver,  the  proportion  will  still  be  true. 

Proof.     If  a  \  h  =  p  :  qy 

a        p 
or  r-  =  ^, 

b         q 

then,  by  multiplying  each  member  by  itself  repeatedly,  we 
shall  have 


-  t 

-Pi 

~  q' 


etc.      etc. 


Hence,  in  general. 


Cor.  If 
then,  from  Th.  IV.  ^ 
and 


rt"  :  b""  =  7J"  :  q\ 
a  :b  =  p  :  q, 
:  a"  ±  b""  =  p""  :  p"  ±  (f 
±  //'  :  b"  =  jy"  ±  f  :  q'\ 
187.  Theorem  VI.      When  any  three  terms  of  n  pro- 
portion are  given,  the  fourth  can  always  be  found  from  the 
theorem  that  the  product  of  the  means  is  equal  to  that  of  the 
extremes. 

We  have  shown  that  whenever 

a  :b  ■:;=■  p  :  q, 
then  aq  =  bp. 


162  RATIO  AND  PROPORTIOK 

Considering  the  different  terms  in  succession  as  unknown 
quantities,  we  find 


h 


aq 

J' 

'  aq 

p  =  -p 

Ip 

^         a 

188,  Theoeem  VII.  If  the  ratio  of  ttvo  quantities  is 
given,  the  relation  bettveen  them  may  he  expressed  hy  an  equa- 
tion of  the  first  degree. 

Proof.     Let  x  and  y  be  the  quantities,  and  let  their  given 

m 
ratio  be  — .     By  the  definition  of  ratio  this  will  mean  that  if 
n         -^ 

nz 
we  multiply  y  by  —  the  result  will  be  x.     That  is, 

711 

X  =  — y. 

Multiplying  by  n, 

my  =nx, 
an  equation  of  the  first  degree. 

Cor.     If  we  know  that  the  ratio  of  two  unknown  quanti- 

m 
ties  is  — ,  we  may  call  one  of  them  mx  and  the  other  nx. 
n 

For  mx  :  nx  =  ~.  (§  178) 


The  Mean  Proportional. 

1 89.  Def.  When  tlie  middle  terms  of  a  propor- 
tion are  equal,  either  of  them  is  called  the  mean  pro- 
portional between  the  extremes. 

The  fact  that  h  is  the  mean  proportional  between  a  and  c 
is  expressed  in  the  form 

a  :  i  =  i  :  c. 

Theorem  T.  then  gives  ¥  =  ac. 


PltOPOllTION.  163 

Extracting  the  square  root  of  both  members,  we  have 

I  =z  Voir. 
Hence 

Theorem  VIII.  The  iuean  proportional  of  two  quanti- 
ties is  equal  to  the  square  root  of  their  product. 

Def.  Three  quantities  are  said  to  be  in  proportion 
when  the  second  is  a  mean  proportional  between  tlie 
first  and  third,  and  the  third  is  then  called  the  third 
proportional. 

EXERCISES. 

What  is  the  mean  i)roportional  between: 
1.   16  and  36?  2.  2  and  8?  3.   a'  and  Z>'? 

4.   a -\- h  '<i\idi  a  —  i"^  5.  ^a  and  9«?        6.   27??./:  and  h^nix. 

T.   What  is  the  tliird  proi)ortional  to  4  and  6? 

There  are  two  ways  of  considering  tbis  question: 

I.  The  third  quantity  must  have  tlie  same  ratio  to  6  that  6 
has  to  4;  that  is,  it  must  be  f  of  6,  which  is  9. 

II.  Because  the  middle  term,  6,  is  the  square  root  of  the 
l)roduct  of  the  first  and  third,  we  must  have 

6  =  V-^  X  third,       or     third  X  4  =  36, 
whence  third  =  9  as  before. 

What  is  the  third  proportional  to: 
8.  m  and  mx"^  9.  2m  and  4w?     10.  m  and  w? 

Problems  in  Ratio  and  Proportion. 

1.  A  poulterer  had  75  chickens  and  geese,  and  there  were 
3  chickens  to  every  2  geese.     How  many  of  each  kind  had  he? 

Solution.  By  §  188  we  may  call  2.^^  the  number  of  his  geese,  and  3aj 
will  then  be  the  number  of  his  chickens.  Therefore  the  total  number 
will  be 

3^'  -f  Zr  =  75, 
which  gives  ,v  =  15. 

Therefore  3.i;  =  45  =  number  of  chickens; 

2,1-  =  30  =  number  of  geese. 

2.  A  drover  liad  5  sheep  to  every  2  cattle,  and  the  number 
of  his  sheep  exceeded  that  of  his  cattle  by  102.  IIoav  many 
had  he  of  each? 


164  BATIa  AND  PROPORTION. 

3.  A  huckster  had  3  apples  to  every  2  potatoes.  Half  the 
number  of  his  apples  added  to  three  times  the  number  of  his 
potatoes  made  960.     How  many  had  he  of  each? 

4.  Divide  198  into  three  parts  proportional  to  the  num- 
bers 2,  3  and  4. 

Kemark.     We  may  call  the  parts  2x,  dx  and  4x. 

5.  Find  three  numbers  proportional  to  1,  2  and  3  whose 
sum  shall  be  192. 

6.  Divide  561  into  four  parts  proportional  to  2,  3,  5  and  7. 

7.  Three  partners,  A,  B  and  C,  had  2,  3  and  6  shares  re- 
spectively. On  dividing  a  year's  profits,  C  had  $600  more 
than  A  and  B  together.  What  was  the  amount  divided  and 
the  share  of  each? 

8.  A  farmer  had  3  cattle  to  every  2  horses,  and  his  horses 
and  cattle  together  were  one  third  his  sheep.  The  whole 
number  was  160,     How  many  had  he  of  each? 

9.  Two  numbers  are  in  the  ratio  3  :  5,  and  if  6  be  taken 
from  each  the  ratio  will  be  5  :  9.     Wliat  are  the  numbers? 

10.  When  a  couple  were  married  their  ages  were  as  5  :  4. 
After  10  years  they  were  as  7  :  6.     What  were  the  ages? 

11.  The  quantity  of  water  in  two  cisterns  was  as  3  :  5. 
After  pouring  12  gallons  from  the  second  into  the  first  they 
were  as  5  :  6.     How  much  water  had  each  at  first? 

12.  Find  two  fractions  whose  ratio  shall  be  2  :  3  and  whose 
sum  shall  be  unity. 

13.  Find  two  fractions  of  which  the  sum  shall  be  unity 
and  the  ratio  x  :  y. 

14.  What  are  those  fractions  of  which  the  sum  is  j-  and 

0 

the  ratio  a  :  b?  .  a''     .       -,       a 

Ans.  -,— T— T9  and 


ab  +  b'  a  -{-b' 

15.  Of  what  two  quantities  is  the  ratio  m,  :  n  and  the  dif- 
ference ^? 

16.  Divide  184  into  three  parts  such  that  the  ratio  of  tlic 
first  to  tlie  second  shall  be  2  :  3,  and  of  the  second  to  the 
third  5:7. 

Suggestion.  If  we  call  x,  y  and  z  the  numbers,  we  may  state  the  pro 
portions  a;  :  y  =  2  :  3, 

2/:s  =  5:7; 
whence,  by  §  181,  8a;  =  %y  and  bz  =  1y. 


PROBLEMS.  165 

17.  Divide  232  into  three  parts  such  that  the  ratio  of  the 
first  to  the  second  shall  be  3  :  4,  and  of  the  first  to  the  third 
2:5. 

18.  A  poulterer  had  5  chickens  to  every  3  ducks  and  1 
goose  to  every  4  ducks,  while  the  entire  number  of  all  was 
175.     How  many  of  each? 

19.  Of  two  trains,  one  went  3  miles  to  the  other^s  4,  and 
the  first  went  50  miles  farther  in  6  hours  than  the  other  did 
in  3  hours.     What  was  the  speed  of  each? 

20.  At  8  o'clock  a  train  left  Washington  for  New  York, 
distance  232  miles.  At  9  o'clock  a  train  left  New  York  for 
Washington,  going  as  far  in  4  hours  as  the  other  did  in  5. 
They  met  at  the  middle  point  of  the  road.  What  was  the 
speed  of  each? 

21.  Of  three  partners.  A,  B  and  C,  A  had  5  shares  and  B 
had  7.  0  sold  A  ^  of  his  stock  and  B  |,  when  he  had  half  as 
much  as  A  and  B  together.  How  many  shares  had  he  at  first  ? 

22.  li  X  :  y  =  b  :  3,  find  the  values  of  the  ratios 

x'.x^y, 
X  :  X  —  y; 
x-\-y:x-y; 
X  —  2y  :  X  -{-  2y; 
y-x  :y; 

y  :2y  -x; 
ax  :  by; 
ax  -\-  by  :  ax  —  by. 

23.  If^±l  =  «,  find-. 

x-y  y 

24.  If  ay  =  bx,  find  ^-^. 

X  tJ 

25.  What  is  the  ratio  of  the  speed  to  the  current  when  it 
takes  a  boat  one  third  longer  to  go  up  stream  than  it  does  to 
go  down? 

26.  When  two  boats  were  steaming  in  the  same  direction, 
one  took  2  minutes  to  get  a  length  ahead  of  the  other. 
When  in  opposite  directions,  they  were  separated  by  a  length 
in  10  seconds.     What  was  the  ratio  of  their  respective  speeds? 

27.  Two  men  ran  a  race  to  a  goal  and  ba  k.  The  swifter 
reached  the  goal  in  40  seconds,  and,  turning  back,  met  the 
slower  2  seconds  later.     What  was  the  ratio  of  their  speeds? 


WQ  PROBLEMS  IN  RATIO  AND  PROPORTION. 

28.  In  a  drove  there  were  2  cattle  to  every  7  sheep,  and 
5  horses  to  every  9  cattle.  What  was  the  ratio  of  the  sheep 
to  the  horses? 

29.  It  takes  one  boat  ^  longer  to  go  up  stream  than  down, 
and  another  \  longer.     What  is  the  ratio  of  their  speeds? 

Suggestion.  First  find  the  ratio  of  tlie  speed  of  each  to  the  velocity 
of  the  current. 

30.  One  cask  is  filled  with  pure  alcohol,  while  another, 
three  times  as  large,  has  equal  parts  of  alcohol  and  water. 
What  is  the  ratio  of  alcohol  to  water  if  the  contents  of  both 
are  mixed? 

31.  A  cask  contained  wine  and  water  in  the  ratio  7:5. 
After  adding  8  gallons  of  wine  and  10  gallons  of  water  the 
ratio  was  5  :  4  and  the  cask  was  full.  How  much  did  the 
cask  hold? 

32.  One  ingot  contains  equal  parts  of  gold  and  silver,  and 
another  has  2  parts  of  gold  to  3  of  silver.  If  I  mix  equal 
weights  from  the  two  ingots,  what  will  be  the  ratio  of  gold 
to  silver  in  the  mixture? 

33.  A  milkman  had  two  cans,  one  containing  milk  and  the 
other  an  equal  quantity  of  water.  He  poured  half  the  milk 
into  the  water,  and  then  poured  half  the  mixture  thus 
formed  back  into  the  milk-can.  What  was  then  the  ratio 
of  milk  to  water  in  the  milk-can? 

34.  The  speed  of  two  trains  was  as  4  to  3,  and  it  took  the 
swifter  train  1  hour  longer  to  make  a  Journey  of  416  miles 
than  the  slower  train  required  for  a  journey  of  273  miles. 
What  was  the  speed  of  each  train? 

35.  If  a  gold  coin  contains  9  parts  of  gold  and  1  of  silver, 
and  a  silver  coin  contains  6  parts  of  silver  and  1  of  copper, 
what  proportions  of  gold,  silver  and  copper  will  an  alloy  of 
equal  weights  of  the  two  coins  contain? 

36.  What  will  be  the  proportions  when  2  parts  of  the  gold 
coin,  as  above,  are  combined  with  1  of  the  silver? 

37.  A  goldsmith  forms  two  alloys  of  equal  weight,  the  one 
composed  of  equal  parts  of  gold  and  silver,  the  other  of  two 
parts  of  gold  to  one  of  silver.  If  gold  is  16  times  as  valuable 
as  silver,  what  is  the  ratio  of  the  values  of  the  two  alloys? 

^^  Teachers  requiring  additional  problems  in  ratio  and  proportion  will  find 
them  in  the  Appendix. 


MEMORANDA  FOR  REVIEW. 


167 


Memoranda   for  Review. 

Ratio. 
Define :  Multiple;   Measure;    Commensurable;  Common 
measure;    Incommensurable;    Ratio  (in  two  ways);   Antece- 
dent;  Consequent;  Terms;  Inverse  ratio. 

Ratio  as  a  quotient. 

i  Integral, 
The  three  classes  of  ratio  I  Fiactional, 

(  Incommensurable. 
Relation  of  direct  and  inverse  ratio. 

^    Result  when  both  terms  are  j  ^|^\^jed^^^  ^>' 


Explain 

and 

Distinguish 


the  same  quantity. 

Result  when  both  terms  are  |  J^^J^ihiTshed  ^^^' 
the  same  quantity. 


Proportion. 
Define :  Proportion;   Terms;   Extremes;    Means;   Inver- 
sion;   Alternation;    Composition;    Division;    Mean    propor- 
tional;  Third  proportional. 

Products  of  Extremes  and  Means,  Theorem 

of;  Prove. 
Inversion,  Theorem  of;  Prove. 
Alternation,  Theorem  of;  Prove. 
Theorems        Composition;  Prove. 
of  Pro-      ■{   Division ;  Prove. 
portion.         Composition  and  division;  Explain. 

Each  term  raised  to  the  same  power;  Th. 
Express  any  term  in  terms  of  the  three  others. 
Express  that  two  quantities  have  a  given  ratio. 
Mean  proportional;  Tlieorem  of» 


CHAPTER  V. 

POWERS  AND  ROOTS. 


Section  I.    Powers  and  Roots  of  Monomials. 

190.  Bef,  The  result  of  taking  a  quantity  A 
n  times  as  a  factor  is  called  the  nth  power  of  A^  and, 
as  already  known,  may  be  written  either 

AAA,  etc.,  n  times,  or  A^. 

Bef,  The  exponent  n  is  called  the  index  of  the 
power. 

Bef,  Involution  is  the  operation  of  finding  the 
powers  of  algebraic  expressions. 

The  operation  of  involution  may  always  be  indicated  by 
the  application  of  the  proper  exponent,  the  expression  to  be 
involved  being  enclosed  in  parentheses. 

Examples.     The  ^^th  power  oi  a -\- h  \^  (a -\-  Vf, 
The  wth  power  of  abc  is  (abcy. 

Involution  of  Monomials. 

191.  Involution  of  Products.  The  nth.  power  of  the 
product  of  several  factors,  a,  I,  c,  may  be  expressed  without 
exponents  as  follows: 

ahc  abc  abc,  etc., 
each  factor  being  repeated  n  times. 

Here  there  will  be  altogether  n  a\,  n  J's  and  n  c's,  so  that, 
using  exponents,  the  whole  power  will  be  a"Z>"c" 

Hence  (abcY  =  aVc\ 

That  is, 

Theorem.  The  potver  of  a  product  is  equal  to  the  pro- 
duct of  the  poivers  of  the  several  factors. 


INVOLUTION  OF  MONOMIALS. 


EXERCISES. 

Form  the  squares,  the  cubes  and  the  nt\i  powers  of: 
1.  mn.  2.  %hp,  3.  ^xy, 

4.  'Zan.  5.  ^r.  6.  bpq. 

192,  Involution    of   Fractions.      Applying    the 

methods  to  fractions,  we  find  that  the  nila.  power  of 

x"- 


r 


For 


XXX         .  .  . 

,  etc.,  n  times 

y  y  y 

xxXy  etc.,  n  times 
yyyy  etc.,  n  times 

a;" 


same 

X    . 
—    IS 


(§^5) 


EXERCISES. 


am 

hn ' 


]^orm  the  squares,  the  cubes  and  the  wth  powers  of: 

2cpq  dpx  abf/z 

4:mqy' 


2. 


dxyz 


^fiinx 


193.  Involutiofi  of  Powers.  Let  it  be  required  to  raise 
the  quantity  a^  to  the  ;^th  power. 

Solution.     The  nth.  power  of  r/"'  is,  by  definition, 

a"^  X  or  X  a^,  etc.,  n  times. 
By  §  36,  the  exponents  of  a  are  all  to  be  added,  and  as 
tlie  exponent  m  is  repeated  n  times,  the  sum 
m  +  m  +  m  +  etc.,  n  times 
is  mn.    Hence  the  result  is  fl^"*",  or,  in  the  language  of  algebra, 
(ary  —a  '"^ 
Hence 

Theorem.  //  any  poioer  of  a  quantity  is  itself  to  he 
raised  to  a  poiver,  the  indices  of  the  potvers  must  he  multiplied 
together. 

Examples.  (a")'  —  a'a^a''  —  r/"; 


170  POWERS  AND  ROOTS. 


EXBRCISES. 

Write  the 

squares 

and  the  cubes  of  the 

■  following  expres- 

sions: 

1.  al)\ 
4.  aVx\ 

2.  2m'n. 
5.   'doVi'x. 

3. 

6. 

fq\ 
bg'^m'^x. 

J.  ^,. 
10.    '"''". 

8    ^^ 
11.   ^^•'^^ 

9. 
12. 

Form  the  wth 

powers  of: 

13.  /^\ 

14.  2m'x. 

15. 

18. 

3pY 

2cr 

19.  ^T. 

».  -r. 

21. 

Wx^' 

Keduce: 

22.  (/r)^ 

23.   (jy''')^ 

24. 

{D^n'xy. 

25.  (2^^=')^ 

26.   {%iil)y. 

27. 

m- 

194.  Algebraic  Signs  of  Potvers.  Since  the  continued 
product  of  any  number  of  j)ositive  factors  is  positive,  all  the 
l^owers  of  a  positive  quantity  are  positive. 

By  §  103,  the  product  of  an  odd  number  of  negative  fac- 
tors is  negative,  and  the  product  of  an  even  number  is  positive. 
Hence 

Theorem.  The  even  poivers  of  negative  quantities  are 
positive,  and  the  odd  poivers  are  negative. 

Examples.     (—  af  =  a"";  (—  ay=  —  a^;  (—  aY=  a*;  etc. 

EXERCISES. 

Find  the  values  of: 

1.   (-3)\  2.  (-  3)^  3.   (-3)\ 

4.   {-ay.  5.  {-ay.  6.   {-ay. 

7.   {-aby.  8.  (-mV)^  9.   {- cn'xy, 

10.  (-l)^  11.  (-1)1  12.  {-ly. 

13.  {-hf^.  14.   (-A)2«  +  i.  15.   {-hf^-K 

16.   (-1)2".  17.   (-1)^"-^  18.   (-1)2"+^ 


ROOTS  OF  MONOMIALS.  171 


Roots  of  Monomials. 

195.  Def.  The  nt\i  root  of  a  quantity  q  is  sucli  a 
number  as,  being  raised  to  the  iit\\  power,  will  pro- 
duce q. 

The  number  n  is  called  the  index  of  the  root. 
The  second  root  is  called  the  square  root. 
The  third  root  is  called  the  cube  root. 
Examples.     3  is  the  4th  root  of  81,  because 

3.  3.3.  3  =  3'  =  81. 
I)ef.  Evolution  is  the  process  of  extracting  roots. 

196.  Sign  of  Evolution^  or  the  Radical  Sign. 

|/,  called  root^  indicates  the  square  root  of  the  quan- 
tity to  which  it  is  prefixed. 

When  any  other  than  the  square  root  is  expressed, 
the  index  of  the  root  is  prefixed  to  the  sign  \/. 

Thus  y,  ^^,  "|/  indicate  the  cube,  the  fourth  and  the  wth 
roots  respectively. 

The  radical  sign  is  commonly  followed  by  a  vin- 
culum extended  over  the  expression  whose 
root  is  indicated. 

Examples.     In  Va-^-h  -\-  x  the  root  applies  to  a  only. 
In   i/a  -{-  b  -\-c  the  root  indicated  is  that  of  a  -\-  h. 
In   Va-\-  b  -\-  c  the  root  indicated  is  that  oi  a  -\-  b  -{-  r. 
But  we  may  use  parentheses  in  place  of  the  vin- 
culum, writing  |/(a  -{-h)  -\-  c^   \/{a  -\-h  -\-c\  etc. 

EXERCISES. 

What  are  the  values  of  the  following  expressions? 


1.  i/7  +  9.     Ans.  4.  2.    1^20  +  5.         3.  i/(3'+  4'^). 

4.  |/(l7  4-8').  5.    Ve .  15  .  1^.     6.  Va\ 

7.  x^ia'-^-^ab-^b'),  8.   Vs!                  9-  Vl^'+^'+S). 

10.  V81.  11.   WW^^^       12.  Vl4^-  13^ 


17^  POWERS  ANt)  MOOTS, 

19*7.  Division  of  Exponents.  Let  us  extract  the  square 
root  of  a\  We  must  find  such  a  quantity  as,  being  multiplied 
by  itself,  will  produce  a\  It  is  evident  that  the  required  quan- 
tity is  a^f  because,  by  the  rule  for  involution  (§191), 

n 

Tlie  square  root  of  dP-  will  be  d\  because  j 

n  M  !!  +  ?! 

n 

In  the  same  way,  the  cube  root  of  cC"  is  a^,  because 

n  n  n 

a'X  a'x  (("'=  (("". 
The  following  theorem  will  now  be  evident: 

Theorem.  The  square  7'oot  of  a  poiver  may  he  expressed 
hy  dividing  its  exponent  ly  2,  tlie  cube  root  ly  dividing  it  by 
3,  and  the  7ith  root  by  dividiiig  it  by  n. 

EXERCISES. 

Find  the  cube  roots  of: 
1.  p\  2.  p\  3.  p\  4.  p^. 

198.  Since  the  even  powers  of  negative  quantities 
are  positive,  it  follows  that  an  even  root  of  a  positive 
quantity  may  be  either  positive  or  negative. 

This  is  expressed  by  the  sign  ± ;  read,  plus  or  minus. 
Examples.      Va""  =  ±  a;    V{a  —  by  =  a  —  b  or  b  —  a. 

199.  If  the  quantity  of  which  the  root  is  to  be  ex- 
tracted is  a  product  of  several  factors,  we  extract  the 
root  of  each  factor  and  take  the  product  of  these 
roots. 

Example.     The  square  foot  of  a^ni^p^  is  ±  a^mp^,  because 
(a^mpy  =  a'nep\  by  §§  191  and  193. 

200.  If  the  quantity  is  a  fraction,  we  extract  the 
roots  of  both  members.  The  root  of  the  fraction  is  then 
the  quotient  of  the  roots. 

Examples,    y  -,-  =  -3;     y  -,  =  ±  ^. 
'   y         y        '    c  c 


FMACTIONAL  EXPONENTS,  175 


EXERCISES. 

Extract  the  square 

roots  of: 

1.  a'm\ 

2.  c'x\ 

3. 

4^2n^4n^ 

'■'?■ 

5      ^^^ 

6. 

7.  (a+bY(a. 

-b)'. 

Express  the  cube  roots  of: 

8aW 

11. 

23h  ^  36n 

Express  the  nth  roots  of: 

12.  x\ 

13.  x^\ 

14. 

cc*"". 

15.  x^\ 

16.  :c^'+^ 

17. 

^n^  +  2n^ 

18.  1". 

20. 

Fractional  Exponents. 

201.  If  we  apply  the  rule  of  §  197  for  division  of  ex- 
ponents so  as  to  express  the  square  root  of  x%  we  shall  have 

Vx'  =  xl 
Because  x  =  x\  we  also  have 

i^x  =  x^  . 
Reversing  the  process  and  applying  the  general  form, 

we  have  ici  X  :r^  =  a:^+i  =  x^  =  x, 

x^  X  x^  =  x^  +  i  =  x\ 

Hence 

A  fractional  exponent  mdtcates  the  extraction  of  a  root. 
If  the  denominator  is  2,  a  square  root  is  indicated;  if  3,  a 
cube  root;  if  n,  an  nth  root. 

A  fractional  exponent  has  therefore  the  same  meaning  as 
the  radical  sign  |/,  and  may  be  used  in  place  of  it. 

It  is  a  mere  question  of  convenience  which  method  of  indicating  the 
root  we  shall  use. 


174  POWEMS  AM)  MOOTS. 

EXERCISES. 

Express  the  following  roots  by  exponents  only, 
1.  V^.  2.  ^{a  -  1)).  3.  V{a  -  b)\ 

4.  V{a  -  bf.  5.  V^.  6.  Va^. 

7.  VI^^\  8.  y^v.  9.  y^. 

cy 
10.  'l/?^.  11.  ^V7^,  12.  V^^iV. 

13.  ;/'^-.      14  y/EEs   15  v^^ 


*    ^    (a-b)y''       ^'  ^  {a-x)y^''  ^  {b-\-x)  {b-x)' 

Express  the  nth  roots  of: 

19.  X.  20.  x\  21.  a:'. 

22.  bmx''.  23.  7c'a;'".  24.  10c'x'\ 

Powers  of  Expressions  with  Fractional 
Exponents. 

^02.  Theorem.   The  pth  poicer  of  the  nth  root  is  equal 
to  the  nth  root  of  the  pth  poiuer. 

Example.  CViy  =  2'  =  4, 

W    =  V64  =  4; 
or,  in  other  words,  the  square  of  the  cube  root  of  8  (that  is, 
the  square  of  2)  is  the  cube  root  of  the  square  of  8  (that  is, 
of  64). 

General  Proof     Let  us  put  x  =  the  nth.  root  of  a. 
The  pth  power  of  this  root  x  wall  then  be  x^.  (1) 

Because  x  is  the  nth  root  of  a, 
x""  =  a. 
Raising  both  sides  of  this  equation  to  the  j^th  power,  w^e 
have 

^np  _  ^p  _  p^Yi  power  of  a. 

The  nth  root  of  the  first  member  is  found  by  diyiding  the 
exponent  by  n,  which  gives 

wth  root  of  j9th  power  =  x^, 


FRACTIONAL  EXPONENTS.  175 

the  same  expression  (1)  just  found  fortlie  pi\\  power  of  the 
nth  Yooi.. 

303,  This  theorem  leads  to  the  following  corollaries: 

CoK.  1.   The  expresmm 

p 
a"" 

1 

mat/  mean  either  the  pth  poiver  of  a"  or  the  nth   root  of  a^\ 

these  quantities  being  equivalent. 

Cor.  2.  The  numerator  of  a  fractional  exponent  ix  the 
index  of  a  power.     The  deno^ninator  is  the  index  of  a  roof. 

Cor.  3.  The  potvers  of  expressions  having  fractio)iaI  ex- 
ponents ?nay  be  for^ned  hy  multiplying  the  exponents  by  the 
index  of  the  poiver. 

EXERCISES. 

Express  the  squares,  the  cubes  and  the  nt\\  powers  of  the 
following  expressions : 

1.  c^.  2.  ch.  3.  cl 

1 
4.  ach  5.  aid  6.  aic''. 


7.  a'y". 

8.  c'l/". 

9.  c'^y^ 

10.  ™*? 
ah 

11.  "*f-. 

1 

7inyi 

^^'  a^b- 

13.  (''+*).:. 

14.  (™  + 

1 

71)'' 
2' 

15    {m-i-71)^ 

{a  —  by  {m  —  n)^  {m  —  7i)» 

204.  Multiplication  of  Fractional  Exponents. 
One  fractional  exponent  may  be  multiplied  by  another 
wlien  a  root  is  to  be  extracted,  according  to  the  rules 
for  the  multiplication  of  fractions. 

The  process  is  the  same  as  that  of  §  193  except  that  the 
exponent  is  fractional  instead  of  entire,  and  roots  are  indicated 
as  well  as  powers. 

Example  1.  Square  root  of  «*  =  («*)^  =  ak 

Ex.  2.  Cube  root  of  m§  =  {ml)\  —  ml. 

Ex.  3.  Square  root  of  {a-\-xYbl  =  \{a-^xYbl\^  —{a-\-x)\U. 


176  POWERS  AND  ROOTS. 

EXERCISES. 

Express  the  square  roots  of: 
1.  xK  2.  ahxi.  3.   {a^h)^x% 

xh  (a  +  b)^  ^m'xi 

(v^  (a  —  ^)3  /i 

Express  the  following  indicated  roots: 
7.   {a^M)l      Ans.  WB^i         8.   (aibi)l       Ans.  «iZ»i 
9.   (7/^3^S)i.  10.   (mhi'^)l 

i  H   i\IL 

11.   \(a^x)^ynlm,  12.   (f/i^)M)i 

13.   ((7"/?")^  14.   (c'^§:r§)i. 

15.   (jo^'^a;'')*.  16.   \m"w"J2 

1  n    m 

Negative  Exponents. 

205.  The  meaning  of  a  negative  exponent  may  be  de- 
fined by  the  formula  of  §  45 : 

rim 

-n    =  «'"-^  (1) 

Let  us  suppose  m  =  3  and  n  =  5.     The  equation  then  be- 
comes 

_  =  «»-=«- 

or,  reducing,  -^  = «-«.  (2) 

We  hence  conclude: 

A  negative  exponent  indicates  the  reciprocal  of  the  corre- 
ifponding  quantity  ivith  a,  positive  exponent. 
Multiplying  (2)  by  «'  gives 

a"  Xa-'=  1. 
Dividing  by  a "" ',  we  conclude 


NEGATIVE  EXP0NE2^'TS.  17t 

Hence 

A  factor  may  he  transferred  from  one  term  of  a  fraction  to 
the  other  if  the  sign  of  its  exjponent  he  changed. 

206,  If,  in  the  formula  (1),  we  suppose  m  =  n,  it  be- 
comes 

a"  =  1. 

Heuco,  because  a  may  be  any  quantity  whatever. 
Any  quantity  with  the  exponent  0  is  equal  to  unity. 

This  result  may  be  made  more  clear  by  succes- 
sive divisions  of  a  power  of  a  by  a.  Every  time 
we  effect  tliis  division,  we  subtract  1  from  the  ex- 
ponent, and  we  may  suppose  this  subtraction  to 
continue  to  negative  values  of  the  exponent.  In 
the  left-hand  members  of  the  equations  in  the  mar- 
gin, the  division  is  effected  symbolically  by  dimin- 
ishing the  exponents;  in  the  right-hand  members 
tlie  result  is  written  out. 

"  EXERCISES. 

Transform  the  following  expressions  by  introducing  nega- 
tive exponents: 

1.  ^-.     Ans.  a'h-\        2    ^' 


a'     = 

aaa 

a'     = 

aa 

a'     = 

a 

a"     = 

1 

a~^  = 

1 

a 

a-''  = 

1 
aa 

etc. 

etc. 

¥'  ^""  ""  •     "•  hh' 

mix"^' 


3.  i^.  4.   i^y    Ans.  a'b'm-'n-'. 


5.  (^y.  6.  0^- 

\pqy  \mnj 

7.  ~.  8.  ^.     Ans.  {x-^h){x~h)-'. 

1     (^^'(^-  +  n)\V'  f  ah'  y 

\  n(m  —  n)  }'  '    Kxy'^zj  ' 

Express  the  following  with  positive  exponents,  reducing 


to  fractions  where  necessary: 


13.  a-\  14.  — ,.  15.  a 


a-'' 


178  POWERS  ajsu  Hoots. 


IG. 

1 

17. 

ab-\ 

18. 

c 

10. 

m' 
m-'' 

20. 

a-'h. 

21. 

22. 

ce'b-''. 

23. 

a-'xh-\ 

24. 

7}1~'^X'^V~\ 

•>■» 

{a-\h)-\ 

2G. 

{a-^x)  (a-^x)-' 

.  27. 

a-^x 

{a^x)-'' 

28. 

{a-\-x)-^ 
{ii-x)-'' 

29. 

a-' 

30. 

1 

^07.  Tlie  rules  of  involution  and  evolution  by  ex- 
ponents apply  to  negative  exj)onents,  regard  being 
paid  to  algebraic  signs. 

-17  11, 

Examples.  — ^  =  -—  =i  a  , 

a~  1 

IF 
or 

(«-')-'  =  {;?)"'=  (^'  -i-y-"'- 

Because  — 2x  — 3  =  + 6^  the  sign  of  tlie  final  exponent 
corresponds  to  the  rnle  of  signs  in  multiplication. 

EXERCISES. 

Free  the  following  exponential  expressions  from  paren- 
theses and  negative  exponents: 

1.   {m~y.     Ans.  -«.         2.   (m-y,  3.   (m-')-'- 

4.   (a¥)-'  5.   {fj-'h)\  6.   (g-'k)-\ 

7.   (/i-^F)-\  8.   (h-'x-')-'-      9-   {h-^x-y. 

10.  (rkj-i)\  11.  {p-kj^)-\    12.  (;^V)-*. 

13.    (;rlvO-*-  14.    (;>-'.r^)'.  15.   {mhj-^)-\ 


THE  BINOMIAL   THEOREM.  179 


Section  II.     Involution  and  Evolution  of 
Polynomials. 

The  Binomial  Tlieoreni. 

208.  Let  us  form  the  successive  powers  of  the  binomial 
1  _^  X.     We  multiply  according  to  the  method  of  §  121; 

Multipliei,  l-\-x 


l  +  x 

+  X     +3? 

(i+^r 

-1+%Z  +  X^ 

1    +x 

l  +  -ix-\-x' 

x  +  2x'  +  x' 

(i+^r 

=  1  +  3x  +  Zx'  +  x' 

1  +  x 

l  +  dx  +  dx"  +  a;' 

X  +  3x'  +  3x'  +  x' 

Multiplier, 


Multiplier, 


(1  +  a-)*  =  1  +  4^:  +  6x'  +  4x'  +  x\ 
It  will  be  seen  that  whenever  we  multiply  one  of  these 
powers  by  1  4-  x,  the  coefficients  of  x,  x^,  etc.,  which  we  add 
to  form  the  next  higher  power,  are  the  same  as  those  of  the 
given  power,  only  those  in  the  lower  line  go  one  place  toward 
the  right.  Thus,  to  form  (1  -\-  x)\  we  took  the  coefficients 
of  (1  +  ^)^  and  wrote  and  added  them  thus: 
Coef.  of  (1  +  x)\        1,     3,     3,     1 

1,     3,     3,     1 

Coef.  of  (1  +  x)\        1,     4,     6,     4,     1 

1,     4,     6,     4,     1 


Again  adding,  1,     5,  10,  10,     5,     1 

And  so  on  to  any  extent. 

It  is  not  necessary  to  write  tbe  numbers  under  each  other  to  add 
them  in  this  way;  we  have  only  to  add  each  number  to  the  one  on  tlic 
left  in  the  same  line  to  form  the  corresponding  number  of  the  line 
below.  Thus  we  can  form  the  coefficients  of  the  successive  powers  of 
X  at  sight  as  in  the  following  table.  The  tirst  figure  in  each  line  is  1; 
the  next  is  the  coefficient  of  ^i  the  third  the  coefficient  of  ic"^,  etc. 


180  POWERS  AND  ROOTS. 

For  practice  let  the  pupil  continue  this  table  to  n  —  10. 


n  =  1; 

coefficients, 

1. 

n  =  2', 

(< 

2,     1. 

71  =  3; 

it 

3,     3,     1. 

w  =  4; 

it 

4,     6,     4, 

1. 

n  =  5; 

a 

5,  10,  10, 

5, 

1. 

n  =  6, 

(< 

6,  15,  20, 

15, 

6, 

etc. 

etc. 

1. 


It  is  evident  that  the  first  quantity  is  always  1,  and  that 
the  next  coefficient  in  each  line,  or  the  coefficient  of  x,  is  n. 
The  third  is  not  evident,  but  is  really  equal  to 

^^>,  («) 

as  will  be  readily  found  by  trial;  because,  beginning  with 
w=3, 

_       3.2      _      4.3      ,_       5.4       , 
3  =  — ,     6  =  -y-,     10  =  -^,     etc. 

The  fourth  number  on  each  line  is 

7i[n  -  l){n  -  2)^  ,^. 

Z  .  o 
Thus,  beginning  as  before  with  the  third  line,  where 

"""    ^_3_^^1      4_iJ^2      10-^-^      etc       (c) 
^-     2.3'     ^~     2.3'     ^^-     2.3'     ^^'''      ^""^ 

209.  The  numbers  given  by  the  f-ormulae  {a),  (b),  (c), 
etc.,  are  called  binomial  coefficients.  In  writing  them, 
we  may  multiply  all  the  denominators  by  the  factor  1  with- 
out changing  them,  so  that  there  will  be  as  many  factors  in 
the  denominator  as  in  the  numerator.  The  fourth  column 
of  coefficients  (6*)  will  then  be  written 

3.2.1      4.3.2     5.4.3 
1.2.3'     1.2.3'    1.2.3'  • 

We  can  find  all  the  binomial  coefficients  of  any  powei 
when  we  know  the  value  of  7i. 

In  every  binomial  coefficient  the  first  factor  in  the  nu- 
merator is  71,  the  second  w  —  1,  etc.,  each  factor  being  less  by 
unity  than  the  preceding  one, 


THE  BINOMIAL   THEOREM,  181 

The  first  factor  in  the  denominator  is  1,  the  second  2,  the 
third  3,  etc. 

The  numerator  and  denominator  of  the  second  coefficient 
will  contain  two  factors,  as  in  {a) ;  of  the  third,  three  factors, 
as  in  {h)  and  (c);  of  the  fourth,  four  factors,  etc. 

EXERCISES. 

1.  Form  all  the  binomial  coefficients  for  the  fiftli  power 
of  1  +  X. 

Ans.   -:^-5;   ^=10;   ^-^  =  —  =  10; 

5.4.3.2  _  5  _       5.4.3.2.1  _ 

1.2.3.4"  1  ~    '  1.2.3.4.5  ~ 
It  will  be  seen  that  after  we  reach  the  middle  of  the  series 
of  coefficients  the  last  factors  begin  to  cancel  the  preceding 
ones,  so  that  the  numbers  repeat  themselves  in  reverse  order. 

2.  Form  all  the  binomial  coefficients  for 

n  =  6;     71  =  7;     n  —  8;     n  =  9. 

310.  The  conclusion  reached  in  the  preceding  section  is 
embodied  in  the  following  equation,  which  should  be  per- 
fectly memorized: 

-,    ,          ,   ^iO^  —  1)    o   ,   n(n  —  l)(n  —  2)     „ 
(l  +  xr  =  l-}-7ix-\-    \,^     \t-4--^ ^^-^3 ^-x' 

"+"  1.2.3.4        "   ^   -h.  .  .  -t-^. 

211.    Def.    When  a  single  expression  is  changed 

into  the  snm   of  a  series  of  terms,  it  is  said  to  be 

developed,  and  the  series  is  called  its  development. 

^  { f2 1 ) 

Example.    The  preceding  series  1  -|-  nx  -j — ^j— r — -  x""  -\- 

etc.,  is  the  development  of  (1  +  xy, 

EXERCISES. 

Develop  the  following  expressions: 

1.   (l  +  ff)*.  2.   (\+ax)'.  3.   (l+^)'- 

4.   (1  +  hj.  5.   (1  +  Vy.  6.  {l+ah')\ 


182  POWERS  AJSD  ROOTS. 

212.  If  the  second  term  of  tlie  binomial  is  nega- 
tive, its  odd  powers  will  be  negative  (§194).  Hence  the 
signs  of  the  alternate  terms  of  the  develojjment  will 
then  be  changed  as  in  the  following  form: 

(1  -  .)»  =  1  -  ..  +  'iif-^'  r  -  ^M  .'  +  ec. 

EXERCISES. 

Develop: 

1.  (1  -  h)\  2.  (1  -  by.  3.  (1  -  by. 

4.   (1  -  ahy.  5.    (1  -  ahy.  G.    [l  -  "'X 

213.  The  developnient  of  any  bmomial  may  be  reduced 
to  the  preceding  form  by  a  transformation.     Let  us  put 

a  =  the  first  term  of  the  binomial; 
b~  the  secoiid  tei'ni; 
WE  the  index  of  tlie  power. 
The  binomial  will  then  be  a  -\-  b.  which  we  may  write 
in  either  of  the  forms 

.{,  +  t)     ,.     .(i  +  ^A). 

The  nth  i)o\ver  of  tlie  iirsi  form  will  be 

By  §  210, 

/        b  Y_  b       n(,  -  1)  b-^       n{n-l){n  -  2)  b^ 

V^aJ  -^^\'^'T72~a-'~^         rr2:3         a'^ 
Multiplying  by  a",  we  luivc 


etc. 


a'-hna 


„    ,,    ,  n(u  —  l)    „    .,,.,   ,   n{n  —  l)(7i  —  2)  ,,3,3,    , 
'^  +  — h — ^  a''-^b'  +  — ^— j — ^---7, Uc''    ^b  -{-  etc. 

We  see  from  this  that,  in  the  develo})ment  of  {a  -\-  by\ 
The  first  term  is  a^'b^  (because  ^"=1). 
The  second  term  is  r/"~^^\  aiul  its  coefficient  is  the  iudvx 
of  the  power. 

In  each  successive  tcrui   the  exponent  of  r/  is  diniiiiislM  d 


THE  BINOMIAL   THEOREM. 


183 


and  that  of  b  increased  by  unity,  while  the  coefficients  are 
the  same  as  in  the  case  of  (1  +  ^')  • 

Remark.  The  work  of  developing  a  power  of  a  binomial 
is  facilitated  by  the  following  arrangement: 

1.  Write  in  one  line  all  the  powers  of  the  first  term,  be- 
ginning wich  the  highest  and  ending  with  the  zero  power,  or 
unity. 

2.  Write  under  them  the  corresponding  powers  of  the 
fiecond  term,  beginning  Avith  the  zero  power,  or  unity,  and 
ending  with  the  highest. 

3.  Under  this,  in  a  third  line,  write  the  binomial  coef- 
ficients. 

4.  Form  the  continued  product  of  each  column  of  three 
factors,  and  by  connecting  with  the  proper  signs  we  shall 
have  the  required  power. 

Example.     Form  the  sixth  power  of  2a  —  3a;^ 


Powers  of  2a,       64a«+  32a« 
"       "  -  3a;2,    1    -     3^a 
Binom.  coef.,  1+6 

(2a  -  3x2)6  z= 


+     16a4     +      8a3      +       4a^     -j-      2a      +1 
4-       9x*     -     27a:«     +     81a;8     _  243a;io   +  729x1 » 
+     15        +20         +15         +6        +1 

64a8-  576a6x2+  2l60a*x*-  4320a3x«+  48ma^x»-  2916axio+  729x»». 
EXERCISES. 


Develop : 
1.   {a  +  by. 


+  .7)' 

7.   {a-i-2xy. 


«.   {a-\-bY. 
5.   (a-a-'y. 

8.   (c-'dh'y. 


10.  (c'-x'y- 

13.  (a-'  +  b-y. 

14.  Write  the  first  five  terms  of  (w  -f-  n 


3.   {a  +  by. 

9.  (»-!)'. 

12. 


KB 


15. 

-  (a  +  bY\ 

16. 

-  (a  -  b)l^ 

17. 

''  {a-  2xy. 

18. 

"  (a  +  3^;)^. 

19. 

"(■+a-- 

184  POWERS  AND  BOOTS. 

Square  Root  of  a  Polynomial. 

214.  Sometimes  the  square  root  of  a  polynomial  can  be 
extracted.  To  extract  a  root,  the  polynomial  must  be  ar- 
ranged according  to  the  powers  of  some  one  symbol.  Sup- 
pose X  to  be  that  symbol.  Then  supposing  the  roof  arranged 
according  to  the  powers  of  x,  calling  71  the  index  of  the  high- 
est power  of  X,  and  calling  a,  h,  c,  etc.,  the  coefficients  of  the 
powers  of  x,  we  shall  have 

Eoot  =  ax''  4-  Z>:i-"-i  +  .  .  .  -f  /. 

Squaring,  we  find  that 

will  be  the  highest  term  of  the  square,  and 

r 

the  lowest  term.     Hence 

A  polynGmial  can  have  no  root  unless  both  its  highest  and 
mvest  terms  are  i)erfect  squares. 

315.  If  the  preceding  condition  is  fulfilled,  ^er/^ajos  the 
polynomial  has  a  root.  If  it  has,  the  root  is  found  by  the 
following 

KuLE.  1.  Arrange  the  polynomial  according  to  ]^owers  of 
that  syinhol  of  which  the  powers  are  most  numerous, 

2.  Extract  the  square  root  of  the  highest  term,  and  subtract 
its  square  from  the  polynomial. 

3.  Fi7id  the  next  term  of  the  root  hy  dividing  the  second 
term  of  the  polynomial  by  double  the  tenn  of  the  root  already 
found. 

4.  Multiply  twice  the  first  term  plus  the  second  term  by  the 
second  term,  and  subtract  the  product. 

5.  Find  each  term  i?i  succession  by  dividing  the  first  term 
of  the  remainder  by  the  first  divisor. 

6.  Multiply  every  new  term  by  tivice  the  part  of  the  root 
already  found  plus  this  term,  and  subtract  the  produ^ct. 

If  there  is  any  root,  the  last  remainder  will  come  out  zero. 

216.  Reason  of  the  Rule.  To  show  the  truth  of  this  rule 
let  us  take  the  polynomial 

tr*  -\-  2(m  +  ^)^'  +  (^^  +  n'^)x'^  —  2mn{m  -\-  n)x  -\-  m^n^.    (a) 
The  square  root  of  the  first  term  is  x^.     Let  us  put  P  for 
the  sum  of  the  remaining  terms,  so  that  the  root  is 

x^-\-P.  (b) 


SQUARE  HOOT  OF  A  POLYNOMIAL.  185 

Then,  by  squaring, 

or,  dropping  x^  fron  each  member, 

2Pa;'  -\-P'  =  2{m  +  7i)x'  +  (ju'  +  n')x'  +  etc.  (c) 

Since  these  two  members  are  to  be  identically  equal,  the 
highest  terms  must  be  equal;  that  is,  the  highest  term  of  2Px^ 
must  be 

2{m  +  n)x\ 
or,  dividing  by  2x^, 

Highest  term  ot  P  =  {m-{-  n)x. 

That  is,  we  find  the  second  term  of  the  root  hy  dividing  the 
second  term  of  the  polynomial  by  twice  the  first  term  of  the 
root.     This  is  No.  3  of  the  rule. 

Next,  by  the  rule,  we  multiply  twice  the  first  term  plus 
the  second  term  by  the  second  term,  and  subtract  the  pro- 
duct.    That  is,  if  we  put,  for  shortness, 
a  =  x^,  the  first  term; 
Z*  =  (m  -|-  n)x,  the  second  term; 
the  quantity  the  rule  tells  us  to  subtract  is 
{2a  +  l))h  =  2aT)  +  h\ 
But,  by  No.  2  of  the  rule,  we  have  already  subtracted 
a^  =  X*,  the  square  of  the  first  term  of  the  root.     Hence  we 
have  subtracted  in  all 

a'  +  2al}  -\-h'={a-\-  h)\ 

which  is  the  square  of  the  sum  of  the  first  two  terms  of  the 
root. 

Reasoning  in  the  same  way,  we  find  that,  at  each  step,  the 
total  quantity  subtracted  from  the  polynomial  is  the  square  of 
the  sum  of  all  the  terms  of  the  root  already  found. 

Hence  if  at  any  step  we  have  zero  for  a  remainder,  it  will 
show  that  the  square  of  the  sum  of  the  terms  of  the  root  is 
equal  to  the  polynomial,  and  therefore  that  the  root  is  really 
the  root  of  the  polynomial. 

The  arrangement  of  the  work  is  shown  in  the  following 
examples,  which  the  student  should  perform,  and  by  which  he 
should  show  that  the  sum  of  the  quantities  subtracted  at  each 
step  is  really  the  square  of  that  part  of  the  root  already  found. 


186  POWERS  AND  ROOTS. 

ErAMPLE  1. 


I  x^-\-{m-\-n)x—mn 


x*-\-%{m-\-n)x^-\-{m''-\-n'^)x^—2mri{m-\-n)x-\-m^7i'' 

X* 


2x^  -]-  {m-\-n)x 

Mult.  {m-}-n)x 

2x''-{-2{m-{-7i)x—mn 

Multiplier,    —fun 


2(m-^7i)x'-{-{7n'-\-n')x' 

2{m^'n)x'-{-(m'^2m7i-{-?i')x' 


—2mnx'' —2mn{'m-^7i)x-\-nfn^ 
—2m)ix^ —2rnn(m-\-?i)x-{-m'^7i'^ 


The  order  of  operations  is  this:  We  find  the  term  x*  of 
the  root  by  extracting  the  root  of  the  highest  term.  We 
write  this  term  three  tiaies:  first  on  the  right,  as  a  part  of 
the  root,  and  then  on  the  left,  as  a  trial  divisor,  and  again  as 
a  multiplier  of  this  divisor. 

Multiplying  x""  X  x'  we  have  x*  to  be  subtracted  from  the 
given  expression.     We  then  bring  down  two  more  terms. 

Forming  x^  ^  x^  =  2x\  the  latter  is  a  trial  divisor,  which, 
divided  into  the  first  term  brought  down,  gives  {m  +  7i)x  as 
the  second  term  of  the  root.  We  write  this  term  three 
times:  first  on  the  right,  then  after  2x\  and  then  under  it- 
self as  a  multiplier. 

Multiplying  the  two  terms  already  found  by  (m  +  n)x,  we 
subtract  the  product  from  the  terms  brought  down,  and  thus 
find  a  second  remainder,  to  which  we  bring  down  the  remain- 
ing terms  of  the  expression. 

Adding  {7n-\-7i)x  to  2x''-{-(m-\-7i)x,  we  have  2a;"+  2{m-{-n)x 
as  the  third  trial  divisor.  Dividing  the  first  term  by  2x\  we 
have  the  term  —  rim  of  the  root,  with  which  we  proceed  as 
before. 


Example  2. 

X*  -  2ax'  +  {a'  -f-  W)x'  -  4:ab'x  +  4J*  \x'-ax  +  2b' 


2x^—  ax 
—  ax 


-  2ax'  -f  (a''  -f  W)x'' 

—  2ax^  -{-  a^x^ 


2x'-  2ax-^2b' 
2b' 


4:b'x'  -  4.ab'x  +  U* 
0  0  0 


SQUARE  BOOT  OF  A  POLYNOMIAL.  187 

EXERCISES. 

Extract  the  square  roots  of: 

1.  a*  -W^^a"  -2a-\~l, 

2.  a*  -  4«'  +  8«  +  4. 

:j.  4:x*  -  I2ax'  +  26a'x'  -  2Aa'x  -\-  16a\ 

4.  x'  -  Gax'  +  16a'x*  -  20a'x'  +  15a'x'  -  6a'x  +  a\ 

b.  2bx*  -  30ax'  +  49«V  -  24:a'x  +  16«\ 

6.  rt*  +  (4^  -  2c)a'  +  {W  -  Uc  +  3c')a^  +  {Uc'  -  2c')a  +  c\ 

7.  «'^>'  +  «V'^  +  b'c'  +  2«^Z>c  -  2ab'c  -  2abc\ 

8.  9:^;^  +  la^c^*  +  10a;' +  4a;  + 1. 

3»  n 

9.  9^:^^"+  12a;'  +  lOa.^  +  4x^  +  1. 

Remark.  The  root  may  be  extracted  by  beginning  with 
the  lowest  term  as  well  as  with  the  highest.  The  pupil 
should  perform  the  above  exercises  in  both  ways.  The  fol- 
lowing is  the  solution  of  Ex.  6,  when  we  begin  with  the  low- 
est term  in  a: 

I  c'-{-{2b-c)a-{-a' 
e   c'+  (4^r  -  2c')a  +  (4Z''  -  4J)c  +  3c')a'  +  {4tb  ^  2c)a'-\-  a^ 


te^i^Zb-    c)a 
{2b—    c)a 


(4^c'  -  2c')a  +  (4^'  -  ^bc  +  3c')a' 
(4.bc^-  2c^)a  +  {W  -  ^hc  +    c')a' 


2c'  -f  (U  -  2c)a  +  a' 


2c'a'-\-{4.b-2c)a'-\-a* 
2c'a'-\-(Ab-2c)a'  +  a* 


0  0  0 


217.  It  is  only  in  special  cases  that  the  root  of  a  poly- 
nomial can  be  extracted .  But  when  it  cannot  be  extracted, 
we  may  continue  the  process  to  any  extent,  though  we  shall 
never  get  zero  for  a  remainder. 

If  the  polynomial  is  arranged  according  to  ascending 
powers  of  the  leading  symbol,  the  degree  of  the  root  in  re- 
spect to  that  symbol  will  go  on  increasing  by  unity  Avith  every 
term  added. 

In  the  contrary  case  the  degree  will  go  on  diminishing, 
tinally  becoming  negative. 


188 


POWMiS  ASD  MOOTS. 


Example.     Extract  the  square  root  of  1 
1        1  -[-  :,  I  1  +  1.,  _  x^x'  +  A^x" 


X. 


5 

la's 


:«:*  + 


"r5T^ 


etc. 


1     1 

2  +  i^ 

X 

+  ¥ 

X  +  ix' 

i+  X 

-  K '  -  K 

-K   -K 

-  ¥'  +  tV*' 

a+  X 

-  i-*^-"  +  iV^" 

+  i^'  -  ijx' 

+  iV-'" 

+  ¥■'  +  tV^'  -  A^^  etc. 

;i+    X 

-  i-^' 

-  -ii^"  +  liV*:'  etc. 

-  ii^"  -  ihx'  etc. 

EXERCISE. 

Extract  the  square  root  of  1  —  x  by  the  aboye  process, 
carrying  the  result  to  x\ 

Square  Roots  of  Numbers. 

218.  The  square  root  of  a  number  may  be  extracted  by 
a  process  similar  to  that  of  extracting  the  square  root  of  a 
])olynomial.  But  there  are  some  points  peculiar  to  a  number 
to  be  understood. 

219.  Niimler  of  Figures  in  the  Root.  By  studying  the 
equations, 

3^  =        9,  or   |/9        =3 

9'  =       81,  or   |/8L       =     9 

10^  =     100,  or   i^'ioO"   =  10 

50'  =  2500,  or   4/2500  =  50; 
Ave  see  that 

If  a  number  has  1  or  2  digits,  its  square  root  has  1  digit; 
If  a  number  has  3  or  4  digits,  its  square  root  has  2  digits: 
If  a  number  has  5  or  6  digits,  its  square  root  has  3  digits; 
etc.  etc. ; 

or,  in  general: 

1.  The  sqtiare  root  has  as  niany  dif/its  as  the  numher  itself 
has  pairs  of  digits,  a  sijiglc  (tUjit  left  over  heiiig  coioifp'l  fs 
a  pair. 


SQUARE  ItUOTS   OF  2s  UMBERS.  189 

2.  The  first  Jigure  (ff  tlte  rout  will  be  the  square  root  of  the 
greatest  perfect  square  contained  in  the  first  figure  or  pair  of 
figures  of  the  numler. 

220.   Case  of  Decimals.     If  we  study  the  equations 
0.9^    -  0.81,       or   |/078l      ==  0.9; 
0.3'    =  0.09,       or   f/(U)9       =0.3; 
0.09^  =.  0.0081,  or   V0.0()81  =  0.09; 
0.03'^  =  0.0009,  or    l/'OOOOO  =  0.03; 
we  see  that: 

1.  The  first  figure  of  the  square  root  of  a  decimal  is  the 
square  root  of  the  greatest  square  contained  in  the  first  pair  of 
figures  of  the  decimal. 

2.  //  the  given  decimal  has  two  or  more  zeros  after  the 
decimal  point,  the  decimal  of  the  root  will  hegin  with  as  many 
zeros  as  the  number  begins  ivith  pairs  of  zeros. 

EXAMPLES    AND     EXERCISES. 

How  do  the  square  roots  of  the  following  decimal  frac- 
tions commence? 


1.  1/0.223.  Ans.  0.4.  l)ccause  0.4  is  the  largest  single  digit 
whose  square  is  contained  in  0.22. 

2.  VO.033.   Ans.  0.1,  because  0.1  is  the  largest  single  digit 
whose  square  is  not  greater  than  0.03. 

3.  Va729.  4.     /a053.  5.    ^0.0092. 
6.    i/0.0005.                  7.    1^000032.  8.    i^'.  00008. 

221.  Extraction  of  the  Square  Root.     Let  us  put 

N  =  tlie  number  whose  root  is  to  be  extracted. 

n  =  the  value  of  the  first  figure  of  the  root. 
This  value  may  be  7i  units,  n  tens,  7i  hundreds,   etc.,  ac- 
cording to  the  place  it  occupies. 

P  =  the  value  of  all  the  figures  after  the  first. 

Q  E  the  value  of  all  after  the  second,  etc. 
Then  we  have,  by  supposition, 

Root  =  Vy  =  ^?  -f  /'. 
etc.         etc. 


190  POWERS  AN2J  HOOTS. 

Squaring,  we  have 

iV^  =  {n  +  Py  =  n'  +  2/iP  +  P'; 
whence 

^nP  -^  P'  =  N  -  n\ 
or 

{%n  -{-  P)  P  =  N  ~  n\ 

BiYiding  by  P,  we  have 

P  =  ^^  ~  '^' 
2n  +  P' 

From  this  equation  we  have  to  find,  by  trial,  the  first 
figure  of  P,  when  we  shall  have  the  first  two  figures  of  the 
root.  Kepeating  the  process,  we  shall  find  a  third  figure,  etc. 
The  general  rule  will  be: 

KuLE.  1.  Separate  the  figures  of  the  member  into  pairs, 
starting  fro7n  the  decimal  point  or  unifs  place. 

2.  The  first  figure  of  the  root  is  the  greatest  number  whose 
square  is  contaified  in  the  first  pair  or  figure  of  the  number. 

3.  Subtract  the  square  of  the  first  figure  of  the  root  from 
the  first  pair  or  figure  of  the  number,  and  bring  down  the 
next  pair. 

4.  Divide  this  result  by  twice  the  first  figure  of  the  root  as 
a  trial  divisor;  add  the  quotie7it  to  the  trial  divisor,  multiply 
the  sum  by  the  qiiotient  and  suUract  from  the  dividend.  For 
the  quotient  must  be  tahen  the  largest  numher  which  will  give 
a  product  less  than  the  dividend. 

5.  After  subtracting  this  product,  bring  dowii  another  pair 
of  figures,  double  the  part  of  the  root  already  found,  and  repeat 
the  process  until  as  many  figures  of  the  root  as  are  wanted  are 
found. 

The  reason  of  this  rule  is  nearly  the  same  as  that  of  the 
rule  for  polynomials.  If  we  call  n,  p,  q,  etc.,  the  values  of 
the  successive  figures  of  the  root,  the  sum  of  the  qiiantities  we 
subtracted  is  the  square  Q>i  n -\- p  ■\- q -\-  etc. 

We  first  point  the  given  number  off  into  pairs  of  digits 
from  the  unit's  place,  as  is  commonly  explained  in  arith- 
metic. We  may  then  liave,  on  the  left,  either  one  figure  or 
two.  The  largest  square  which  it  contains  is  the  first  figure 
of  the  root.  We  subtract  the  square  of  this  figure,  and  use 
double  the  root  figure  as  a  trial  divisor.     This  gives  1  as  the 


SQUARE  ROOTS  OF  NUMBERS.  191 

second  figure,  which  we  affix  to  the  trial  divisor  and  also  use 
as  a  multiplier.  Subtracting  the  product  again,  we  use  double 
the  part  of  the  root  as  a  second  trial  divisor,  and  so  on. 

After  getting  a  certain  number  of  figures,  we  may  find 
nearly  as  many  more  without  changing  the  divisor.  We  may 
tlien  cut  off  one  figure  from  the  divisor  for  every  one  we  add 
to  the  root. 

As  an  example,  let  us  find  the  square  root  of  26890348. 

26890348  |  5185.5904196 
25 


101)  189 
1    101 


1028)  8803 
8  8224 


10365)  57948 
5  51825 


10370.5)  6123.00 
5  5185.25 


10371.09)  937.7500 
9  933.3981 


10i3|7!l.|l|8)  4.3519 
4.1485 


2034 
1037 


997 
933 

~64 
62 


EXERCISES. 

Extract  the  square  roots  of: 
1.  2193.         2.  46084.        3.  293462. 
4.  89189219.      5.  82.6292.      6.  41693. 


192  POWERIS  AND  BOOTS. 

Section  III.    Operations  upon  Irrational 
Expressions. 


Definitions. 

222.  Def.  A  rational  expression  is  one  in  which 
the  only  indicated  operations  are  those  of  addition, 
subtraction,  multiplication  or  division. 

All  the  operations  we  have  hitherto  considered,  except  the  extraction 
of  roots,  have  led  to  rational  expressions. 

Def.  A  perfect  square^  cube  or  jith  power  is  an 
expression  of  which  the  square  root,  cube  root  or  nth. 
root  can  be  extracted. 

223.  Def.  An  expression  which  requires  the  ex- 
traction of  a  root  is  called  irrational. 

Example.     Irratiomil  expressions  are 

or,  in  the  language  of  exponents, 

a^,         {a  +  h)h,         27i 
Def.    Expressions  irrational  in  form  are  called  ir- 
reducible when  incapable  of  being  expressed  without 
the  radical  sign. 

Remark.     Only  irreducible   expressions  are  ])roperly  called  irra- 
tional. 

Example.     The  expressions 

though  irrational  in  form  are  not  so  in  reality,  because  they 
are  equal  to  «  -f  ^  and  6  respectively,  which  are  rational. 

Def.  A  surd  is  an  indicated  root  which  cannot  l)e 
extracted  exactly. 

Def.  A  quadratic  surd  is  a  surd  with  an  indi- 
cated square  root. 

Example.  The  expression  a  -j-  ^  \^  is  irrational,  and  tlie 
Burd  is  VXi  which  is  a  quadratic  surd. 


AG0RE0A2I0N  OF  SIMILAR  8U1W8.  191-^ 

DeJ\  Irrational  terms  are  similar  when  they  con^ 
tain  identical  surds. 

Examples.  The  terms  -/SO,  7  l/30,  {x  +  y)  V^O  are 
similar,  because  the  quantity  under  the  radical  sign  is  30  in 
each. 

The  terms  {a  +  i)  Vx  -\-  y,  ^  Vx  -\-  y,  m  Vx  -j-  y  are 
similar. 

Agj^regation  of  Similar  Surds. 

234.  Irrational  terms  may  be  aggregated  by  the  rales  of 
§§  28  and  30,  the  surds  being  treated  as  if  they  were  single 
symbols.     Hence 

When  similar  irrational  terms  are  connected  hy  the  sig7is 
-\-  or  — ,  the  coefficients  of  the  similar  surds  may  be  aggre- 
gated, and  the  surd  itself  affixed  to  their  sum. 

Example.     The  sum 

ax/{^-^y)-i  y  (^  +  ^)  +  3  V{^  +  y) 

may  be  transformed  into  {a  —  b  -\-  ^)  ]/(x  -\-  y). 

EXERCISES. 

Reduce  the  following  expressions  to  the  smallest  number 
of  terms: 

1.  dVa-  5ab  V2  +  6  |/«  +  Hab  V2. 

2.  6«f  -  7b'^  +  8ah  +  9bk 

3.  (a'by  +  {b'cc)h  -  aa\  -  bbl   Ans.  {a  -  b)bh  -{-{b-  a)aK 

4.  {c'x)h  +  {b'yY^  -  bxh  -  cy^. 

5.  {a  +  b)  Vx-\-  (a  -  b)  Vx. 

6.  a(xh  +  yh)  +  b{xh  -  yi). 

7.  a(xi  -  yh)  +  b{yi  -  zh)  +  c{zh  -  x\). 

8.  {d  +  c)  sT^^J^y  -  [cl  -  c)  V^~^y. 

9.  7  x/x  -f  y  -f  8  Vx  -  y  -  5  Vx -}- y  -  7  Vx^^. 

10.  i  Vm  -j-  n  —  ^  Vm  —  u  -\- ^  Vm  -f-  7i  —  ^  V'm  —  n, 

11.  i{a  +  b){  Vx-\-  Vy)  -  i{a  -  b){  V^  -  Vy). 

12.  (m  4-  n){ai  -  M)  -f  {m.  -  n){ah  +  M). 

13.  (m-{-n-2p)(ai^bh-  c^)^{7n  -2n-^ p){ai -M~-hd) 

-f-  (-  2m  +  n  +  p)(--  ai  +  M  -f  c^). 


194  fowehs  and  hoots. 

Multiplication  of  Surds. 

225.  Irrational  exjoressions  may  sometimes  be  trans- 
formed so  as  to  have  different  expressions  under  the  radical 
sign  by  the  method  of  §  197,  applying  the  following  theorem: 

Theorem.     A  root  of  the  product  of  several  factors  is 
■equal  to  the  product  of  their  roots. 
In  the  language  of  algebra, 

\ahcd,  etc.  =  \^  Wb  Vc"  V^,  etc. 
1111 
=  aH^'d^d^',  etc. 
Proof.     By  raising  the  members  of  this  equation  to  the 
jiQi  power,  we  shall  get  the  same  leeult,  namely, 
a  X  b  X  c  K  d,  etc. 

Example.  V3Q  =  1^4  l/9, 

ivhich,  by  extracting  the  roots,  becomes 

6  =  2.3, 
ii  true  equation.. 

EXERCISES 

Prove  the  truth  of  the  following  equations  by  extracting 
the  roots  in  both  members: 

VI  1^49  =  VTM. 

VI  V2b  --=  VIM. 
VI  VM  =  VIM. 
1/94/25  =  VdM. 

226.  Ap^j)lication  of  the  Theorem  to  the  Multiplicatiori  of 
Surds.  The  preceding  theorem  gives  the  following  rule  for 
finding  the  product  of  two  or  more  surds  with  equal  indices: 

Rule.  Form  the  product  of  the  quantities  under  the  radi- 
cal sign,  and  ivrite  the  similar  indicated  root  of  the  product. 
If  this  root  can  he  exactly  extracted,  the  product  is  rational, 

EXERCISES. 

Express  each  of  the  following  products  by  a  single  surd, 
or  without  surds. 

1.  Vx  X  \^.  Ans.    Vxy. 

2.  VbX\^.  3.     \7ixVbX\'c. 
4.    \^i  X  Vp  X  V"/. 


MULTIPLICATION  OF  8U1WS.  195 

6.  {a  +  by  {a  -  b)K  Ans.  {a'  -  by, 

7.  {x  +  yy{x^  y)K 

8.  '/ic  4-  «2/  X  Vx  —  ay. 

9.  |/a''+  1  X  Va-\-l  X  l/a-  1. 


10.    Vm'x  V^^J  X  V'm  +*7i. 


11.  /5  X  '/S  X  V«+^. 

12.  Va  X  Vb  X  Va  +  b  X  V^  -  b. 

13.  i/«J  X  /^  X  Vck      Ans.  ^?^>c. 

14.  {2m7iyx(27npYx{:npy. 

15.  2%i^  X  2^?i^  X  2imK 

16.  2K«  +  by  X  2i{a  -^  J)*  X  (a -{-  b)K 

17.  (aJc)^  X  (bcdy  X  (cda)i  X  {dab)K 

18.  a  Vx  Vy  X  b  Vx  Vy  X  c  Vx  V~y, 

19.  a(«  +  by  X{a-  by  4-  b(a  +  &)*  X  (tt  -  J)i 

Ans.  {a-{-b){a''-by. 

20.  a  Vx~^  X  Vx  —  y  ^b  Vx -]- y  X  Vx  —  y. 

21.  V^niTi  X  V2mx  X  VnxX  Vy- 

22.  |/6  X  VT^  X  I^IO.  23.    «*^1  +  -J- 

24.  ;^iif^  -  ~-]\  25.    |/(^+^X  i/(m+^i)t 

26.  V2'x  V2  X  'VTe.  27.  V(«+  J)  X  W{a  -  b). 


28. 


V'^  X  |/-^.  29.   l/^  X  V- 

^    n  in  ^    aA-  b        ^ 


1 


m  ^   a-\-  b  {a-\-  by 

30.  r  -y-  X  r  yj.  31.  r  -—^  x  y  — — y. 

A  A^  a  —  b  a-{-  b 


32. 


/;4^  X  /f -^  X  j/y-fr  X  / 


A  -|-  ^  li  —  k  h  -\-k  h  —  k 


Factoring  Irrational  Expressions. 

227.  Conmrsely,  if  the  quantity  under  the  radical 
sign  can  be  factored,  we  may  express  its  root  as  the 
product  of  the  roots  of  its  factors.       (Comp.  191,  199.) 


t96  POWERS  A^n  JWOTS. 

EXERCISES. 

Factor: 
1.   {aljc}i.  Ads.  ciiMc^.  2.   {mnp)^. 

3.    Va""  —  ah.  Alls.    Va  \'a  —  b. 


4.    \^m^  —  mx.  5.    \d'—'b'' 


6.  ^a'  -  b\  7.    Va'  -  a^Z>^ 

8.  1/55.  9.     V66. 

10.  (?//^' -  4m  V)^.  11.   (a'b-aby. 

12.  (a:^;^' +  «V  +  «':r)i  Ans.   r^ia;%^+ «:?; -f  r;^)^. 

13.  {a'm'-a'm'+a'niy.  14.   (P'^  _  .^^^-^^^ 

228.  When  the  quantity  under  the  radical  sign 
is  factored,  one  of  the  factors  may  be  a  perfect  power. 
We  may  then  extract  the  root  of  this  power,  and 
affix  the  surd  root  of  the  other  factor  to  it. 

In  the  problems  of  this  and  the  next  three  articles  the  ob- 
ject is  to  remove  as  many  factors  as  possible  from  under  the 
radical  sign. 

EXERCISES. 

Factor: 

1.  V¥b.        Ans.    Va'  Vb  =  aVb. 

2.  Vrnri^.  3.    Vnfn^.         Ans.  mn  Vn. 


4.  Vm'n'x\  5.  Va'bc\ 

6,  (g'rY)l  7.  {4.a'x)h. 

8.  4/56.         Ans.  2  VU.  9.  |^'27. 

10.  V36^.  11.  V2  Vn. 


12.    Vx\a-\-b).  13.    Va'b-a'b\ 

14.    |/Zt4-  Ubx  +  b'^x,         Ans.   {a  -\-b)Vx. 

Here  the  quantity  under  the  radical  sign  is  equal  to 
(^2  4.  2ah  +  W)x  =  {a-[-  hfx. 

In  questions  of  this  class  the  beginner  is  apt  to  divide  an  expression 
like  ^^a  -\-  b  -\-  c  into  Va  -\-  Vb  -\-  Vc,  wliich  is  wrong.  The  square  root 
of  tlie  sum  of  several  quantities  cannot  be  reduced  except  by  factoring. 
Hence  such  an  expression  as  the  square  root  of  a -{-b-\-c  is  not  reduci- 
ble, but  must  stand  as  it  is. 


FACTOlUJSiJ  IRRATIONAL  EXPRhJSlSlOJSS.  197 


15.   (m^  —  'im^n  +  m7i^)h.  16.    ym*  —  2m^n  -\-  m^n^ 


17.    VAa^  -  Hacy  +  4:c'y.         18.    Va'  -  a'b\ 


19.    V(fb'-b\  20.  4/4wV-T6?. 

n.  (m'-tmy.  22.  (^^^  +  w^)i 

23.  (a'  -  2a'  +  r?)i  24.  (a?^''  +  4«;^'  +  4:any. 

25.  (w^  +  4^i^  +  47z^)i 

229.  When  a  quantity  to  be  factored  by  the  pre- 
ceding method  is  affected  by  a  fractional  exjionent, 
this  exponent  may  be  divided  into  an  integer  (positive 
or  negative)  and  a  fraction,  and  the  quantity  factored 
accordingly. 

Exj^mples.         ai  =  a^+i=  a^ .  ah 

a-i  =  a-^+i  =  a^^ .  a^  =  — . 

a 

EXERCISES. 

Factor: 

1.  ml  ■  2.  chnh 

3.  chul  •  4.  8i         Ans.  16  V2, 

5.  24i  6.    t'24^. 

7.  ax^  -h  hx^  -f  cxl         Ans.  (a  +  bx  +  cx'^yxi. 

8.  2mi  +  3«//?i  —  4.7i^ml 

9.  x-i-{-xK  Ans.   fl  +  -j  xK 

10.   «-^a:i-«*a;--*.  11.  a*  +  2«§  +  3^§. 

12.  aixi-\-aixl  13.  8i^  +  8^a;i 

230.  The  preceding  method  may  be  applied  to  any  other 
indicated  roots  as  well  as  to  square  roots. 

Example  1.    Factor  'VT6  =  16i 

We  have  16^  =  8^  X  2^  =  2 .  21. 

Ex.  2.    Factor  W~a^  +  ^Va^. 

Solutio7i.  a~l-{-  a^=  a~^.  a\  -{-  aal  =  (—  ■+-<?) a». 

\a        i 


198  POWERS  AND  ROOTS, 


EXERCISES. 


Factor: 

1.  24i  2.  54§.  3.  32S.  4.   161. 


5.  c§  -f  6-3.  6.  c§  +  ct.  7.   Va'^*'  -  h\ 

231.  The  preceding  system  of  removing  factors  from^ 
under  the  radical  sign  may  be  applied  to  fractions,  applying 
the  principles  of  §§  200  and  228. 


EXERCISES. 

Factor : 


125*  '  m' 


4. 


x^       x^                                        mx^  —  nxi 
o.    —r-\ J-.  y. 5—7 — . 

x'  +  a:^\4 


10.  .    ,   ,      3 

a"  4-  «'/ 

Reduce  the  following  fractions  to  their  lowest  terms: 

ax^  4-  ex  .        a  -\-C'xi            myi  —  ny^ 

11.     : .         AnS. ~.     xZ. — ; '—rl 

ax^  —ex  a  —  cxi            my^  -\-  nyi 

aVg  +  h  va 

15.     -— z=.  lb.    -^ 7T^. 

^12  9^  -  g^h 


17.  — :; .    Ans. 


h  —  m'  '  ^fi 


m 


18   i-^lz:^.  19     ^  +  ^* 


a-}-u;   *  '  am  —  IrnS" 


MULTIPLICATION  OF  8UBDS.  199 


21. 
23. 


r((l  +  m)  +  h{l  +  ^^)^ 
a(l  +  7i)  -  b{l  H-  7^)^• 


Va'  -  26^=*  +  «'^ 


"  {c  +  A)^  -  1* 

22.   ^  +  ^;. 
3-^3 

„.     («-^)* 

l/aV-2«2;'^  +  ^'^'  *   («'-ic^)^* 


Multiplication  of  Surds  of  Different  Degrees. 

232.  To  multii^ly  surds  of  different  degrees  they  must 
first  be  reduced  to  surds  of  the  same  degree. 

Pkoblem  I.  To  reduce  surds  of  different  degrees  to  surds 
of  the  same  degree. 

KuLE.  Express  the  indicated  roots  hy  fractional  exponents, 
and  reduce  tlte  expo7ie7its  to  their  L.  CD. 

EXERCISES. 

Reduce  to  surds  of  equal  degree: 

1.  Va,     V¥,    Wc.        Ans.  a^,  U,     cl 

2.  V^,    Wx,    Wx.                   3.  Vm,    Vm,    Vm. 
4.   Wm\   \''iJi\   Wm.                5.  (a  -  b)i,   (a  +  b)i. 

333.  Problem.  IT.    To  express  a  product  of  surds  of 
different  degrees  as  a  simple  surd. 

Rule.  Reduce  the  surds  to  the  same  degree,  and  express 
the  product  hy  the  fundamental  theorems  (§§  191,  202,  225). 

Example.     Express  as  a  single  surd  the  product  of  the 
square  root  oi  a  by  the  cube  root  of  b. 
am  =  am  =  {a')i  {by  (§  203)  =  {aW)i  (§  226)  =  W¥b\ 

The  general  principle  applied  in  this  case  is  this: 

When  ice  have  to  form  a  product  of  several  factors  affected 
with  fractional  exponents  having  a  common  deiiominator, 

We  may  remove  the  denominator  from  all  the  exponents 
and  write  it  outside  the  parenthesized  product  of  the  factors. 

That  is,  we  always  have 

T    1/    2  1 

a">V^»  ....=:  {a''W(f .  .  .  .)". 


200  POWERS  ANJJ  Room. 

EXERCISES. 

Express  with  single  surds  the  products; 


1.     V2xW'd. 

Ans. 

72i 

2.  3*2i 

3.  am. 
5.  a^Uxk 

7.  mxk 

9.  2^{x-y)i. 
11.  2m. 

1    1 
4.  xp^y''. 

6.  2ai         Ans.  (4a)i. 

8.  32*«i^i 

10.  a^(x -^  y)\  (:c  -  y)h, 
12.  3H^5i 

13.  t 

14    (^^'  +  2/)^ 
{x  -  yY 

234t.   Co7'ollary.      Because  a  rational   quantity  may  be 

considered  as  affected  with  the  exponent  — ,  the  preceding 

method  can  be  applied  to  reduce  the  product  o<*  q  rational 
quantity  and  a  surd  to  the  form  of  a  surd. 
Example  1.     Eeduce  2h^  to  a  single  index. 

2m  =  2m  =  (2T)^  =  {8by  =Wsb\ 

Ex.  2.     dabkr:h  =  dia^x^  =  (SWb'xy  =  'VU^a'b'x 
Ex.  3.     a^h  Vm  =  a  +  VWi. 

EXERCISES. 

Reduce  to  single  surds : 

1.  2Vb.  2.  3  Vx.  3.  bax^. 

4.  m  Vn.  ■  5.  (m  +  n)  Vm  —  n, 

6.  a  Va.  7.  {a  +  b)  Va-{-b. 

8.   (m  -  n)(m  ~  n)-i.  9.     ^^  +  ^ 


10.  «(a  +  J)i  11.  m{m  —  7i)'i. 


1    m 


^^^    «^^  ^g      xy'z^ 


h  fk^m^ 

X    x^      •  ax^y\ 

y'  y'  '  cp^q^' 


MULTIPLICATION  OF  lliii ALTON AL  Ji]XL^TiE88L0NS.     201 

235.  Irrational  expressions  may  be  multiplied  by  multi- 
plying each  term  of  the  multiplier  into  each  term  of  the  mul- 
tiplicand and  taking  the  sum  of  the  products. 

Example.     Develop  the  product  {a  -[-  b  Vx){g  -\-h  Vy). 
a  +  bV^ 

ff  +  ^i'^y 

ag  +  bg  Vx 

ah  Vy  +  ^^^  ^^y 
Product  =  ag  -{-  bg  Vx  -\-  ah  Vy  +  bh  Vxy, 

EXERCISES. 

Multiply: 
1.  {a-c  V^){a  +  b  Vy).  2.   (a  +  Vy){a  -  Vy), 

3.   {a  +  n  Vx){b  —  m  Vy).  4.  ai{aib  —  a^x). 

5.   (am  -\-  n  Va  —  x){n  —  m  Va  —  x). 


Ans.  mnx  -j-  {rC  —  am"^)  Va  —  x. 
6.   {:m-nVy){m  +  nVy).  7.  {a -\- c  Vx){a  -  c  Vy). 

8.   {a' -\- c  V^^^^a){c  ^  a  V^^'^^a). 


9.   {a  +  Va'  -  x'){a  -  Va'  -  x'). 


10.  (m  -  ^m'-%n'){m  +  Vm'-  2n'). 

11.  (m  +  V7n'  -  l)(m  -  |/m^  -  1). 

12.  (V^-{.Vb)(Va-  Vb). 

13.  (c+  i^+  |/^)(c  -  i/^-  Vy). 

14.  (a*  +  M  +  c*)(«^*  -bi+  ci){ai  -\-bi  -  ci)(ai  -  §i  -  c*). 

15.  ai[l-{a-l)i][l-]-2(a-l)i]. 

16.  (  I^M^  +  Vx  -  a){  Vx-{-a  —  Vx  —  a). 

17.  {m  +  l/^+  l/^)(m  -  Vx^  Vy). 

18.  (a:  +  «/  -  |/J-  Vy){l  +  l/^+  i/^). 


202  POWERS  AND  ROOTS. 

\Vx-a        J\  Vx  -a        ) 
21.   \{i)t  +  1)^  +  '^A  [(^^^'  +  1)^  ~  H- 

236.  Rationalizing  Fractions.  The  quotient  of  two 
surds  may  be  expressed  as  a  fraction  with  a  rational  numer- 
ator or  a  rational  denominator  by  multiplying  both  terms  by 
the  proper  multiplier. 

Example.     By  multiplying  both  terms   of    —^  by  4^6, 

we  have 

1/5"  _  6 

and  the  numerator  is  rational. 

In  the  same  way,  multiplying  by  Vl ,  we  have  ^ 

VJ_  _  v^ 

Vl~       7* 

EXERCISES. 

Express  the  following  fractions  with  rational  denominators: 

1    -^.  2.  -^ 

'    i^d*  '2  1^2 


3.    i^  4    -l^^_±i 

^     Vm  ^     Vx'-l 


^    \m  —  n]  '  d  c 

9.  "4.  10.  i. 

ni  al 


RATIONALIZING  FRACTIONS.  203 

237.  If  one  term  of  tlie  fraction  contains  a  quad- 
ratic surd,  it  may  be  rationalized  by  multiplying  by 
tlie  term  itself  with  the  sign  of  the  surd  changed. 

Example.     To  rationahze  the  denominator  in 

Vr 

we  multiply  both  terms  hj  P  -\-  Q  V R.      The  numerator 
becomes 

QR-{-P  Vr. 
The  denominator  becomes 

{P-Q  VR){P  -\-QVR)  =  P'-  Q'R. 
So  we  have  _ 

V~R       ^  QR^P  \^R 

P-QVR~    P'~g'R    ' 

EXERCISES. 

Eeduce  the  following  fractions  to  others  with  rational 
denominators: 

^    a-^-h  Vx  ^  a  —  c  Vy 

a  —  h  Vx  a  -\~  c  Vy 

^         Vh  ,  Vx 


Vb  a  -\-  c  Vx 


f,         Vm  -\-  n  c!    ci-\-  x^ 

a  —  Vm  —  n  a  —  x^ 

7     ^+  ^  3     Wn-\-   \/{m-{-n) 

VA-  Vb  '     Vm-  \/{m  +  7i) 

^     V^-  V(a-x)^  ^^^              1 


Vx-\-  \/{a  —  x)  m  -\-  Vm^  —  a^ 

11               1  12  (^^^  +  nY  ~  {m  -  ny 

'  m  —  Vrrf  —  a''  '  {m  +  n)^  +  (?^  —  n^ 

13.           ^-y_.  14.       ^ 


x-\-y-2Vxy  ^-^'^'^ 


204  POWERS  AND  ROOTS. 

Irrational  Factors. 

338.  By  introducing  surds  many  expressions  can  be 
factored  which  are  prime  Avhen  only  rational  factors  are  con- 
sidered. The  following  theorem  may  be  applied  for  tliis 
purpose: 

Theorem.  The  differeyice  of  any  ttvo  quayitities  is  equal  to 
the  product  of  the  sum  and  difference  of  their  square  roots. 

Proof.    If  a  and  h  are  the  quantities,  we  shall  have 

a  -  h  =  {ai  -  l)^){a^  ^  h^), 
as  can  be  proved  by  multiplying,  or  by  §  123. 


Factor: 

EXERCISEa 

1.  a  —  X. 

2. 

a 

-1. 

3.  a"  -  bx. 

4.   16  -  3. 

5. 

1 
a 

_  1 

c  ' 

6.  x'  -{a  +  by 

7.  {x  -  cy 

X 

8. 

x'- 

■  {P  +  q)' 

9.  {x-cf- 

-i{p- 

-q)' 

10. 

X 

a 

.1 

c  ' 

11.  1  -  ia 

+  ^). 

12. 

m"- 

m  - 

-  n^      m  —  n 

■  n        m  -{-  n' 

13.  x'-^.ax 

+  4.a' 

-I. 

14. 

x"- 

-2x-6. 

15.  x'~2ax-i- a'- {p-\-q).     16.  2a'- 4:{p  -  q). 

239.  Irrational  Square  Roots.  When  a  trinomial  con- 
sists of  two  positive  terms  with  twice  the  square  root  of  their 
product,  its  square  root  may  be  found  by  the  method  of  §  131. 

EXERCISES. 

Find  the  irrational  square  roots  of  the  following  expres- 
sions: 

Note.     A  few  of  the  roots  are  rational. 

1.  a  —  2  Vab  +  b.       Ans.  l/«  —  Vb. 

2.  a-^2Vab-^b. 

3.  «  +  -  +  2.  Ans.  ai  +  \, 

a  a* 


IRRATIOJS'AL  F AC  TOMS.  205 

'  a 

6.  a'  -  2  +  4.  7.  «^"  -  2  +  i. 

8.  9  +  5  -  6  y^.  ^*   ^  + 


1^4+3-  +  ^-  11-   ^  +  2^  +  16- 

12.  ?^-?i  +  !:i.  13.  ^-2  +  ^' 


16      10^25'  '   c 


a 


14. 4  H .  15. 6-4 . 

c  a  n  m 

16.  i»  +  y-2H r—. 


17.  (x^y)-^{x-y)-^2V¥^^\ 

18.  ic-l"  2:^+1.  19.  ai-'Za^-^rl. 


20.  «i  —  2^  +  «i  21. 


4  2     "^  ~4  * 


22.  a;§  +  2^:  +  .-ci  23.  x^  —  x -\-  -x%. 

24.  mi-2  +  m-i  25.   (a;  +  «) +  2(:c  +  a)i+l. 

26.  4(^  —  «)  +  4wi(a;  -  ^)*  +  m\ 

27.  (a:  -  <^)  -  2«i(:?;  --  a^  +  a. 

28.  l-^+2c|/l--  +  c». 

29.  «'Z>' +  4  +  4^-''^-*. 

340.  Square  Roots  of  Irrational  Binomials.  If  we  have 
such  a  binomial  as  «  +  Vh,  its  square  root  may  be  expressed 
in  the  form 

^a-^Vh, 

'  in  which  a  surd  is  itself  under  the  radical  sign. 

If  this  root  can  be  so  reduced  that  no  snrd  shall  be  under 
the  radical  sign,  it  is  said  to  be  reducible;  otherwise  it  is 
irreducible. 


206  i"0  WEllS  AMJ   HOOTS. 

Problem.  To  lind  whether  the  square  root  of  an  irra- 
tional binomial  is  reducible,  and  when  it  is  to  express  it  in 
the  reduced  form. 

Solutioii.     Let  us  suppose 

Squaring,  a-\-Vb  =  x-\-y-\-^  Vxy. 

This  equation  may  be  satisfied  by  putting 
x-\-  y  =  a, 

h 
xy  =  j. 


By  putting  ^a  —  Vb  =  Vx  —  Vy,  we  get  the  same  ex- 
pressions for  X  and  y. 

Therefore,  if  we  can  find  tivo  rational  quantities,  x  and  y, 
such  that  their  su7n  shall  he  equal  to  the  ratiojial  terrn  of  the 
binomial  and  their  product  to  one  fourth  the  square  of  the  surd 
term,, 

Then  the  root  is  the  sum  or  difference  of  the  square  roots 
of  these  quantities. 

Example  1.  Vb^%V<6  =  ^5+  VYl 

We  have  to  find  two  numbers  of  which  the  sum  shall  be  5 
and  the  product  24  -t-  4  =  6.  Such  numbers  are  2  and  3. 
Therefore 

^5  +  2  1/6=  |/3  +  \% 

and  ^5-2  V^ :^  Vd-  1^2.    ' 

Remark.  We  see  that  when  we  express  the  binomial  in  the  form 
a  ±  2  V'c,  a  will  be  the  sum  and  c  the  product  of  the  required  numbers 

EXERCISES. 

Express  the  square  roots  of  ; 

1.  7  +  2  V^,  2.  7  -  4  |/3. 

3.  7  _  2  1/6.  4.  9  -  1/80. 

5.  3  _^  |/8.  6.  6  +  4  i^. 

7.  6  +  V^,  8.  7  +  ^^■ 

9.  8  +  |/60.  10.  14-8  \^. 


TO   COMPLETE  TEE  SQUARE.  207 


11. 

15  -  6  y^. 

,  12. 
14. 

17  +  4  Vl5. 

13. 

2m  +  2  V'm'*  -  1. 

m  -1-  \/m'  —  1. 

15. 

2  -  2  |/1  -  m\ 

16. 

18. 

1  -  fl  -  m\ 

17. 

2«  +  2  Vcc"  -  ^^ 

a  -  Va"  -  b\ 

19. 

^'^  +  2a  Vb'  -  d\ 

20. 

4.a-^2V^a'-b\ 

21. 

2a  -  b-  2  Va'-ab. 

22. 

4:X-2i/4:X'-a.\ 

I 


To  Complete  the  Square. 

241.  If  one  term  of  a  binomial  is  a  perfe'ct  square, 
such  a  term  can  always  be  added  to  the  binomial  that 
the  trinomial  thus  formed  shall  be  a  perfect  square. 

This  operation  is  called  completing  the  square. 

Proof.  Call  a  the  square  root  of  the  term  which  is  a  perfect 
scjuare,  which  term  we  suppose  the  first,  and  call  w  the  other 
term-,  so  that  the  given  binomial  shall  be 

a^  -\-  m. 

Add  to  this  binomial  the  term  — ^,  and  it  will  become 

4rr 

This  is  a  perfect  square,  namely,  the  square  of 

,    m 

H 

that  is,  „=  +  ,„  +  g==(„  +  |Ly. 

Hence  the  following 

Rule.  Add  to  the  binomial  the  square  of  the  second  term 
divided  by  four  times  the  first  term. 

Example.     Wluit  term  must  be  added  to  the  expression 
x^  —  4:ax 
to  make  it  a  perfect  square? 

The  rule  gives  for  the  term  to  be  added 
{-4:axy       .  , 

4:X^ 

Therefore  the  required  perfect  square  is 
x"  -  4:ax  +  4^"  =  (x  -  2ay, 


208  F0WER8  AND  ROOTS. 

EXERCISES. 

Complete  the  square  in  each  of  the  following  expressions, 
and  extract  the  root  of  the  completed  square: 

1.  a\-  2ab.  2.  «' +  4:ax. 

3.  4«'  -  Sax.  4.  4:a'  +  4:a'x. 

6.  a'  -\-b.  6.  a'  -  h. 

7.  a'  -  4tt^  8.  a'x'  +  a'x. 

342.  In  the  preceding  article,  by   adding  4«*  to  the 
binomial  x^  —  Aax,  we  formed  the  equation 
x""  —  4:ax  +  4a*  =  {x  —  2a)  ^ 
By  transposing  the  added  term,  we  have 
x^  —  4:ax  =  (x  —  2ay  —  4a^ 

The  original  binomial  is  now  expressed  as  the  difference 
of  two  squares.  Therefore  the  above  process  is  a  solution  of 
the  problem 

Having  a  Imomial  of  wliich  one  Irr^'^  is  a  perfect  square, 
to  express  it  as  a  differ ejice  of  two  sqtKn^  -. 

EXERCISES. 

Express  the  following  binomials  as  differences  of  two 
squares: 

Ans.  {a  -hY  -  l\ 

3.  x^ -^^ax. 


1. 

a^  -  lah. 

2. 

a"  -  4a5. 

4. 

rr'  +  2ax. 

6. 

x'  _  2x 

a^        a' 

8. 

Aa'x'  -  Ah'x, 

10. 

mV  +  2, 

12. 

;..-■ 

14. 

:c'        X 
4a'  +  «' 

5. 

a\v^  —  a^x. 

7. 

a'    '   ■^* 

9. 

w'a;'  -  1. 

1. 

1+. 

13.  a'x'  -  Wx. 


15.  .'+;^. 


MEMOBAJ^BA    FOE  REVIEW. 
Memoranda  for  Eeview. 


'2m 


Involution.  - 


Fotuers  cmd  Roots  of  Mono7nials. 

Define:  Power;  Square;  Cube;  nih.  Power;  Index  of 
power;  7ii\\  Koot;  Index  of  root;  Square  root; 
Cube  root;  Evolution. 

Of  products;  Theorem;  Proof. 
Of  fractions;  Theorem;  Proof. 
Of  powers;  Theorem;  Proof. 
Algebraic  sign  of  powers;  Theorem. 

Sign  of  evolution;  Index;  Vinculum. 

Division  of  exponents;  Theorem. 

Sign  of  even  root. 

Of  products;  Eule;  Proof. 

Of  fractions;  Rule;  Proof. 

Indicate  root;  Explain. 
Theorem  of  power  and  root  (§  202). 
Significance  of  terms  of  the  fractional  expo- 
nent; Explain. 

Powers )     „  .        -.      .       ^      , .       , 

P     ,      j-  of  expressions  havnig  fractional  ex- 
ponents; How  formed. 

Meaning;  Explain  application  of  preceding 
rules  to  negative  exponents. 


Evolution. 


Fractional 
Exponents 


Negative    ( 
Exponents.  ] 


Binomial 
Theorem. 


Square 

Root  of 

Polynomial 

or  Number. 


Powers  and  Roots  of  Polynomials. 

Powers  of  1  +  :r;  How  formed. 
Expression  for  coefficients. 
Define  binomial  coefficients. 
When  one  term  is  negative. 
Powers  of  a  +  &• 

Arrangement  of  terms. 
Criterion  that  a  root  is  possible. 
^    Eule  for  polynomial;  Explain. 

When  exact  root  cannot  be  extracted.      ' 
Square  root  of  numbers;  Rule;  Explain. 


210 


POWERS  AND   IIOOIS. 


Surd  Fac- 
tors and 
Products. 


Irrational  Expressiotis. 

Define  :  Rational;  Irrational;  Perfect  power;  Irreducible; 
Surd;  Quadratic  surd;  Similar  surds. 

Aggregation  of  Similar  surds;  Rule. 

Theorem  of  roots;  Proof. 

Products  )     „        -,      „  T 

-p    ;  I      surds  01  same  degree. 

When  one  factor  is  a  perfect  power;  Rule. 
Case  of  fractions. 

Reduction  of  surds  to  common  degree;  Rule. 
Product  of  surds  of  same  degree;  Rule. 
Of  irrational  polynomials. 

Rationalizing  Irrational  fractions;  Rule. 

To  factor  any  binomial;  Theorem. 
Irrational  square  roots. 
Square  roots  of  binomial  surds;    When   re- 
ducible; Expression  for  root. 

To  complete  the  square;  Rule;  Result. 


Irrational 

Factors 
and  Roots. 


CHAPTER  V. 

QUADRATIC  EQUATIONS. 


Section  I.    Purp:  Quadratic  and  Other 
Equations. 

243.  Def.  A  quadratic  equation  is  one  which, 
when  cleared  of  fractions,  contains  the  second  and  no 
higher  power  of  the  unknown  quantity. 

Remark.  A  quadratic  equation  is  also  called  an  equa- 
tion of  the  second  degree. 

Def.  A  pure  quadratic  equation  is  one  which 
contains  no  power  of  the  unknown  quantity  except 
the  second. 

Def.  A  complete  quadratic  equation  is  one  which 
contains  both  the  first  and  second  powers  of  the  un- 
known quantity. 

Pure  Quadratic  Equations. 

244.  If,  on  clearing  of  fractions,  arranging  according  to 
powers  of  x  and  transposing,  we  put 

xi  =  the  coefficient  of  x^^ 
B  =  the  terms  not  containing  x, 
a  pure  quadratic  equation  will  reduce  to 

Ax'  =  B. 
Dividing  by  ^,  we  have 

.      B 

x=--. 

Extracting  the  square  root  of  both  members,, 

'B  _      & 

T~  "^  Ar 


212  QUADRATIC  EQUATIONS, 

!<545.  Positive  and  Negative  Roots.  Since  the  square 
root  of  a  quantity  may  be  either  jjositive  or  negative,  it  fol- 
lows that  when  we  have  an  equation  such  as 

x^  =  a 
and  extract  the  square  root,  we  may  have  either 

X  =  -\-  ai 
or  X  =  —  a^. 

Hence  there  are  two  roots  to  every  such  equation,  the  one 
positive  and  the  other  negative.  We  express  this  pair  of 
roots  by  writing 

X  =  ±  aif 
the  expression  ±  #  meaning  either  -\-  a^  or  —  aK 

Remark,  It  miglit  seem  that  shice  the  square  root  of  x^  is  either 
-\-xoT  —  X,  we  should  write 

±  X  =  ±  a^, 
having  the  four  equations       x  =       a^, 

X  =  —  a^, 

—  X  =  -\-  a^, 

—  a;  =  —  a^. 

But  the  first  and  fourth  of  these  equations  give  identical  values  of  x 
by  simply  changing  their  signs,  and  so  do  the  second  and  third;  hence 
more  than  two  of  the  equations  are  unnecessary. 

Example.     Solve  the  equation 

X  —  na  _  nx  —  h 
X  —  a        X  —  b' 

Clearing  of  fractions,  we  ha^e 

{x  —  b)(x  —  na)  =  (a:  —  a)(nx  —  b)y 
or  x^  —  bx  —  nax  +  ^«^  =  ^i^^  —  i^ct^  —  bx-\-  ah. 

Removing  equal  terms, 

x^  +  nah  =  nx"^  +  ab. 
Transposing,  nx"^  —  x^  =  nab  —  ab, 

or  (n  —  l)x''  —  (n  —  l)ab. 

Hence  x"^  =  ab. 

Extracting  root,  x  =  ±  Vab. 


FUUE  QUADHATIC  EQUATIONS.  213 

246.    Studying  the  preceding  process,  we   see   that  a 
pure  equation  is  solved  by  the  following 
Rule.     1.    Clear  the  equation  of  fractions. 

2.  Transpose  all  terms  containing  x^  as  a  factor  to  one 
member;  tJiose  not  containiyig  x  to  the  other. 

3.  Divide  hy  the  aggregated  coefficient  of  the  square  of  the 
imknown  quantity. 

4.  Extract  the  square  root  of  loth  memhers. 

EXERCISES. 

Solve  the  following  equations,  regarding  x,  y  or  z  as  the 
unknown  quantity: 

1.  x""  =  a.  2.  x'  =  c\ 

3.  dx"  =  27.  4.  2x'  -  98  =  0. 

5.  3x'  -  36  =  0.  6.  7x'  -  4:  =  0. 

7.   (a  -  h)x'  =  c.  8.   {a  -  b)a''  =  a'  -  h\ 

9.  mx'-  m''^  nx'-^  n^=  0.       10.   (x  +  5){x  -  5)  =  11. 

11.  (x-  a'){x  -  c')  +  (a'  +  c')x  =  3a' c\ 

12.  ax^  —  b-i-c  =  0. 

13.  (x  -  a){x  -  b)  +  (:r  +  a){x  +  b)  =  a\ 

^.  a       ff  ~  x'  7,2 


16. 


(x'  +  ay 

10      _        8^  1 

x'-i-2-  6x^  -  32-  ^^'  '"^  +  ^  =  ^''^- 

m 
18.  ^.r  4-  _-  :::=  ^^:^,.  jg^  x{a-x)  -  x(a  +  :z,-)  =  0. 

22.   a  :  .^'  =  a;  :  ^.  23.   «  :  ^.-c  =  c^  :  «. 

24.  x'-^-a  -.x"  —  a=p-[-q  :p  —  q, 

25.  (:^-  +  «  +  ^)  (^  -  a  +  ^)  +  (a;  4-  «  -  J)  (:^  _  «  _  J)  =  0. 

27.  ^^a-^2b  :x^a-2b  =  b-2a-^2x:b-{-2a--2x. 
Reduce  by  composition  and  division  (§  185). 


214  QVABliAl'IC  EqUATIOJSS. 


Pure  Equations  of  any  Degree. 

347.  Def.  An  equation  wMcli,  when  cleared  of 
fractions,  contains  only  the  Tith  power  of  the  un- 
known quantity  is  called  a  pure  equation  of  the  nth 
degree. 

A  pure  equation  of  the  third  degree  is  called  a  pure 
cubic  equation. 

One  of  the  fourth  degree  is  called  a  pure  biqua- 
dratic equation. 

Def.     A  pure  equation  reduced  to  the  form 

x""  ^  a  (1) 

is  called  a  binomial  equation. 

24:8.  Solution  of  a  Binomial  Equation. 

1.  When  the  exponent  is  a  wliole  nuniber.  If  we  extract 
the  ^^th  root  of  both  members  of  the  equation  (1),  these 
roots  will,  by  Axiom  V. ,  still  be  equal.     The  /^th  root  of  x"^ 

1 
being  x,  and  that  of  a  being  a'\  we  have 

X  —  a"", 
and  the  equation  is  polved. 

2.  When  tlie  exponent  is  fractional.     Let  the  equation  be 

x:^  =  a. 
Raising  both  members  to  the  /?th  power,  we  have 

x"^  —  a^. 
Extracting  the  mth  root, 


X  =  a"^. 
If  the  numerator  of  the  exponent  is  unity,  the  equation 
will  take  the  form  i 

x""  =  a. 

By  raising  it  to  the  ^th  power, 

X  =  a^. 
Hence  the  binomial  equation  always  admits  of  solution  by 
forming  powers,  extracting  roots,  or  both. 


PURE  EQUATIONS  OF  ANY  DEGREE.  215 

EXERCISES. 

Find  the  values  of  x  in  the  following  equations: 

2.  -r  =  a, 

X 


1. 

a 

¥  =  "• 

3. 

Xi 

5. 

8       X' 

X      27* 

7. 

x'     ~       X      ' 

9. 

x^      ei 

11. 

V7nx'  -  ¥  =  X. 

Square  both  members. 

12. 

{x  -  h)^  =  mK 

-iA 

/^2       1       7,2\1    _                   ^ 

4. 


x^  —  c      x^  —  a 


6.  ^  =  »K 

X^ 

x^       m 

O.     —  —r. 

a       oc^ 


10.    \/{x^  -  a")  =  c. 


13.  {(x^  -  d)^  =  qh. 


(x'  -  ny 

249.  Problems  leading  to  Binomial  Equations, 

1.  Find  two  numbers  one  of  which  is  three  times  the 
other,  and  the  sum  of  whose  squares  is  90. 

2.  Find  two  numbers  of  which  one  is  twice  the  other,  and 
of  which  the  product  is  48. 

3.  Two  numbers  are  required  whose  ratio  is  2  :  3  and  the 
difference  of  whose  squares  is  61:|^. 

Note  §  188,  Cor. 

4.  Two  numbers  are  required  whose  ratio  is  3:5  and 
the  difference  of  whose  cubes  is  264G. 

5.  Find  three  numbers  of  which  the  sccoiul  i,s  i  \vi(;c  the 
first,  the  third  tAvice  the  second,  and  tlie  sum  of  the  ,<(]iiarc& 
378. 

6.  Find  a  number  such  that  if  4  l)e  added  lo  i{,  mikI  sub- 
tracted from  it  the  product  of  the  sum  uud  difference  shall 
be  273. 

7.  Of  what  two  numbers  is  one  twice  the  olher,  and  the 
difference  of  the  squares  60? 


216  QUADRATIC  EQUATIONS. 

8.  Find  two  numbers  wliose  ratio  is  2  :  3  and  the  sum  of 
whose  cubes  is  945. 

9.  Find  two  numbers  whose  ratio  is  3  :  4  and  the  square 
of  whose  sum  is  392. 

10.  What  number  multiplied  by  its  own  square  produces 
1331? 

11.  What  number  multiplied  by  its  own  square  root  pro- 
duces 729? 

12.  Find  tw^o  numbers  in  the  ratio  2  :  5  tlie  difference  of 
whose  squares  is  greater  than  the  square  of  their  difference 
by  216. 

13.  Find  two  numbers  in  the  ratio  m  :  n  whose  product 
is  rt^ 

14.  Find  two  numbers  in  the  ratio  m  :  n  the  difference  of 
whose  squares  is  greater  by  m^  —  7i^  than  the  square  of  their 
difference. 

15.  What  number  multiplied  by  its  own  square  root  pro- 
duces 48  V^? 

16.  Of  what  number  is  the  product  of  the  square  and  cube 
roots  243? 

17.  Find  three  numbers  in  the  ratio  1:2:3  the  sum  of 
whose  squares  shall  be  350. 


Section  II.     Complete  Quadratic  Equations. 

250.  Normal  Form.  Every  complete  quadratic 
equation  can,  by  clearing  of  fractions  and  transpos- 
ing, be  reduced  to  the  form 

ax"  +  J^  -f  c  =  0, 
in  which  a,  h  and  c  may  represent  any  numbers  or 
algebraic  expressions  which  do  not  contain  the  un- 
known quantity  x. 

Def.     The  form 

ax'  +  hx-{-c  =  0  (1) 

is  called  the  normal  form  of  the  quadratic  equation. 

The  quantities  a^  h  and  c  are  called  coefficients  of 
the  equation. 


SOLUTION  OF  qUADRATIC  EQUATIONS.  217 

251.  General  Form.     If  we  divide  all  the  terms  by  a, 
the  equation  (1)  will  become 

x'  -\--x-\--  =  0. 
a         a 

Kow  let  us  put,  for  brevity, 

h  c 

^       a  ^      a 

The  last  equation  will  then  be  written 

x'  ^px^q  =  0.  (2) 

Def.     The  form 

x^  -\-px-\-  q  =  0 
is  called  the  general  form  of  a  quadratic  equation, 
because  it  is  a  form  to  which  every  such  equation  may 
l)e  reduced. 

Solution  of  a  Complete  Quadratic  Equation. 

252.  The  Equation  m  its  Ge7ieral  Form. 

AVe  first  transpose  q,  obtaining  from  the  general  equation  (2) 

x^  -\-  px  —  —  q. 
By  §  241  the  first  member  may  be  made  a  perfect  square 

by  adding  ~. 

Adding  this  quantity  to  both  members,  we  have 

^  -{-P^-h-^-^'j-Q- 
The  first  member  of  the  equation  is  now  a  perfect  square. 
Extracting  the  square  roots  of  both  members,  we  have 

Transposing  ^,  we  obtain  a  value  of  x  which  may  be  put 
in  either  of  the  several  forms, 


p       Vf- 

-i, 

2-^          3 

f 

or  x  =  i{  —  p  ±  yp"  —  ^q), 

and  the  equation  is  solved. 


218  QUADRATIC  EQUATIONS. 

^53.  The  Two  Roots.  Since  the  square  root  in  the  ex- 
pression for  X  may  be  either  positive  or  negative,  there  will 
be  two  roots  to  every  quadratic  equation,  the  one  formed 
from  the  positive  and  the  other  from  the  negative  surd.  If 
we  distinguish  these  roots  as  x^  and  x^,  their  values  will  be 


_  -p-  4/(7/  -  4g) 

^«  -  ^z 

Example  1.     Solve  the  equation 

X  ^^      _        9 

X  —  S      X  -\-  '6 
Clearing  of  fractions, 

x'  +  Sx-  2a;'+  6x  =  -  2(x'  -  9) 
=  -2x'  -{- 18. 
Transposing  and  arranging, 

x'  +  9x  =  18. 
Completing  the  square, 

2    .    n      1    81        1Q    1    81        153 
^  +9:f-f-  =  18  +  -^-  =  -j- 

9         .    V153 


(3) 


Extracting  the  root,         x  -{-  —  =  ± 
whence  x  = 


2  "  -"       2    ' 

-  9  ±  VTEs 


2 

Ex.  2.  -x'  -\-Sx-l  =  0. 

The  coefficient  of  x"^  being  negative,  we  must,  in  ord^r  to 
form  a  perfect  square,  change  the  sign  of  each  term.  The 
equation  then  becomes 

x'-3x-\-l  =  0, 


or 
Completing 

x'- 
square, 

x'- 

-3x 

-3x-^ 

9 
4 

_  9 
~  4 

1. 
-  1  = 

5 

Extracting 

root, 

X  — 

3 

2 

=  ± 

2  ' 

whence 

X 

_  3  ±  1/5 

2 


SOLUTION  OF  QUADRATIC  EQUATIONS.  219 


EXERCISES. 

Solve: 

1.  x'  -  ix  =  3.  2.  x"  -^x  =  0. 

x  —  ^_'^x  —  ^  x  —  ^_x-\-l 


x-\-%       x-\-ll  a;  +  2      2a; -8 

5.  x"  -  2x  =  3.  6.   2;'  -  5a:  =  14. 

7.  2{x  -  2){x  -S)  =  (x-  l){x  +  2)  -  16. 

8.  {2x  +  2){x  4-  3)  -  (a;  -  l)(a;  +  2)  =  5. 

9.  x'  -j-ax-{-b  =  0.  10.  a;'  +  «a:  -  ^'  =  0. 
11.  a:'  -  «a:  +  Z>  =  0.                  12.  x'  -  ax  -  b  =  0. 

254.  TAe  Equation  in  its  Normal  Form.      If  in  the 
equation  (1), 

ax?  -^hx  -\-  c  =■  ^, 

we  transpose  c  and  multiply  the  equation  by  «,  we  obtain  the 
equation 

d'x^  +  abx  =  —  ac. 

To  make  the  first  member  a  perfect  square,  we  add  —  to 
each  member  (§  241),  giving 

a^x^  +  alx  4-  —  = ac. 

Extracting  the  square  root  of  both  members,  we  have 

ax-\-  -  —  -  |/(&^  —  4«c). 
We  then  obtain,  by  transposing  and  dividing, 

-'b±  \/{y  -  4:ac) 


-  =  -£±i^(*'-^«^) 


2a 
Hence  the  two  roots  are 


_-b-j-{b'  -  4.ac)i 
x^  — 

and  x„  =  — 


2a 

{If  -  ^acy 


2a 


(4) 


We  can  always  find  the  roots  of  a  niven  quadratic  equation  by  sub- 
stituting tlie  coefficients  in  tlie  preceding  expression  for  x.  But  the 
student  is  advised  to  solve  eacli  separate  equation  by  the  process  just 
iriven,  wliich  is  embodied  iu  the  followina:  rule. 


220  Q UADRA TIV  KQ L  A  TWyS. 

255.  Kulp:.     1.  Reduce  the  equation  to  its  normal  or  its 

general  form,  as  may  be  most  cofivenient. 

2.  Transpo.se  the  terms  which  do  not  coidain  x  to  the  sec- 
ond member. 

3.  If  the  coefficient  of  x^  is  unity,  add  one  fourth  the 
square  of  the  coefficient  of  x  to  both  members  of  the  equation 
and  extract  the  square  root. 

4.  If  the  coefficient  of  x^  is  not  unity,  either  divide  by  if 
so  as  to  reduce  it  to  unity,  or  multiply  or  divide  all  the  terms 
by  such  a  factor  or  divisor  that  the  term  in  x"  shall  become  a 
perfect  square. 

5.  But  if  the  term  in  x^  is  already  a  perfect  square,  no 
multiplication  or  division  need  be  performed. 

6.  Complete  the  square  by  the  rule  of  §  241,  and  extract 
the  square  root. 

Example.     Solve  the  equation 

7  =  2x. 

X  —  4: 

Clearing  of  fractions  and  transposing,  we  find  the  equation 
to  become 

2x'  -    9x  +  1  =  0. 

X  2,         '  4:X^—  18:k  =  —  2. 

92       81 
Adding  x  ~  IT  ^^  ®^^^^  member  of  the  equation,  we  have 

4^   -  I8.7;  +  --  ^  -  -  2  =  --. 


ll.itracting  the  square  root, 
2x 


9  _  V73  _  1/73^ 
2  ~  ^    4'  ~      2  ' 


whence  we  find 

x 

9  ±  V73 
~         4 

So  the  two  roots 

are 

^. 

=  i(9  +  4/73), 

.?•„ 

=  i(9  -  1  73). 

SOL  UTJOS   01^    Q  VA  DBA  TIC  EQ  UA  TIONS.  2^1 

EXERCISES. 

Solve  the  equations: 

1.  x''^Jix  =  ilc\ 
Steps  of  Solutio7i. 

h  _{h^  +  k'^f 
^  +  3"  -         2       ' 

_  -h±  jh''  +  A;^)* 
X  -  2  ' 

or  .  »i  = ^-y-^ , 

-Ti-  {M  +  ^2)^ 
x,= — ^ 

2.  a:''  +  2Aa;  =  F.  3.  ^'^  +  2^:r  +  ¥  =  0. 
4.  x''  +  4m^  -  7z  =  0.  5.  2;'  -  2ax  =  3a\ 

6.  a;'  -  4:ax  +  4«'  =  0.  7.   m'2;'  +  2/irz;  =  p. 

8.  wa;'  -  m'^  =  4m'.  9.  «a;'  -  2bx  =  W, 

10.  4  _  2-  -  3  =  0.  11.   |-^  +  6-  =:  3m. 

x^ 
12.   — ,-  —  mx  =  n.  13.  m'a;^  —  2m'ic  +  m*  =  1. 

14.  4:cx'  -  {a  +  b)x-\-c  =  0. 

15.  (a  +  b)x'  +  («  -  ^')a;  =  a'  -  b\ 

X  —  a 
m  — 


16    ^~  ^  I  ^  —  g_Q  17    _^^_±_^  =  _  1 

*^'  —  C.T  —  a       2*  *         .:?;  +  « 

m  H ' — 

X  —  a 

18    _1-  =  1+1  +  1  19    ^_?.  =  i!4.i 

a; +  5      a^S^x"  3       X       S^  x' 

30.  i  -  -  =.  i  +  «.  31.  i+      1      =L+L. 

m        X       n       X  X       X  -\-  a      a       2a 

22    _^4_-J_=_15_      as    ^  +  1    ,    ■^-1^4a.--l 

a;  +  2  "^  it-  +  4      a;  +  6'  a;  +  2  "^  a;  -  2      2x  -  V 

24.  a'  +  5'  +  cc'  =  l-2aJa;. 

25.  {a  -  xy  +  (.'T  -  hy  =  (a  -  hY. 


222  QUADRATIC  EQUATIONB. 

26.  (^  -\-¥  -\-  x'  =  2{ax  -i-bx-i-  ah). 
^^_    {x  -  ay  +  (:,  -  ^^  _  a^  -f  h' 


(.^  _  af  -{x-  by  ~  a'  -  ¥' 

28.  ^  +  -  =  ^.  29.  a; +  -  =  3. 

X       'Z  ^   X 

30.  X =  — -.  31.  ic =1. 

X        4  X 

32.  a;  +  -  =  — .  33.  X —  c. 

X       n  X 

In  the  following  equations  find  the  value  of  y  in  terms 
of  X,  and  of  x  in  terms  of  y\ 

34.  x""  +  xy  +  I/'  -  c'  =  0. 

-y  ±  Vic'  -  Zy'            -x±  VI^^^' 
Ans.  X  -  — ^ =^;   y  = ^ . 

35.  x""  -  xy  -^  if  -  h'  =  0.       36.  y''~dxy-\-'2x''-x-y  =  c. 
37.  a;^  -  4^y  +  4^^  =  0.  38.  x"  +  7ia;?/  +  ny'  =  0. 

39.  alf^  -h  «-Ja:?/  +  i'y'  =  0.      40.   x'  -  Q—xy  +  9-^.y'^  =  0. 

256.  Problems  leadmg  to  Quadratic  Equations. 

1.  Find  two  numbers  of  which  the  sum  is  32  and  the  pro- 
duct 231. 

2.  Find  two  numbers   of  which   the   sum  is  ])  and  the 
product  q. 

3.  Find  two  numbers  of  which  the  sum  is  40  and  the  pro- 
duct 48  times  the  difference. 

4.  Of  what  two  numbers  is  the  difference  10  and  the  pro- 
duct 375? 

5.  Find  those  two  numbers  whose  sum  is  38  and  the  sum 
of  whose  squares  is  724 

6.  Find  those  two  numbers  whose  sum  is  m  and  the  sum 
of  whose  squares  is  n". 

7.  Find  those  two  numbers  whose  difference  is  m  and  the 
difference  of  whose  squares  is  n^. 

8.  Divide  the  number  26  into  two  parts  whose  product 
added  to  the  sum  of  their  squares  shall  be  556. 

9.  Divide  20  into  two  such  parts  that  the  square  of  their 
(difference  shall  be  equal  to  their  product. 


PROBLEMS.  223 

10.  Divide  12  into  two  such  parts  that  the  sum  of  their 
squares  shall  be  4  times  their  product. 

11.  Can  there  be  two  unequal  numbers  the  square  of 
whose  difference  is  equal  to  the  difference  of  their  squares? 

12.  A  drover  bought  a  certain  number  of  sheep  for  $663. 
Had  he  bought  them  for  II  apiece  less  he  would  have  got  48 
more  sheep.     How  many  did  he  buy? 

Note.  If  he  bought  x  sheep,  each  sheep  must  have  cost  him  — 
dollars, 

13.  When  tea  costs  50  cents  a  pound  more  than  coffee  you 
<an  buy  20  pounds  more  of  coffee  than  of  tea  for  $7.50. 
What  is  the  price  of  each? 

Note.  After  writing  the  equation  simplify  it  by  dividing  out  all 
<()inmon  factors  from  its  terms. 

14.  An  almoner  divided  $10  equally  between  two  families. 
Tliere  was  one  member  more  in  the  second  family  than  in  the 
first,  in  consequence  of  which  each  member  of  the  second  got 
25  cents  less  than  each  member  of  tlie  first.  What  was  the 
number  of  each  family? 

15.  Find  two  numbers  in  the  ratio  2  :  3  such  that  three 
times  the  lesser  added  to  the  square  of  the  greater  shall  make 
99.     (§  188,  Cor.) 

16.  The  length  of  a  rectangular  lot  exceeds  its  breadth  by 
55  feet,  and  it  contains  1564  square  feet.  What  are  its  length 
nnd  breadth? 

17.  A  rectangular  field  takes  120  yards  of  fence  to  sur- 
round it,  and  contains  3375  square  yards.  What  are  its 
length  and  breadth? 

18.  A  lot  120  feet  by  25  feet  is  to  be  surrounded  by  a 
border  of  uniform  breadth  containing  250  square  feet.  What 
is  the  breadth  of  tlie  border?  (Express  the  result  in  decimals 
of  an  inch.) 

19.  The  outer  walls  of  a  house  20  feet  x  40  feet  on  the 
outside  cover  224  square  feet.     What  is  tlieir  thickness? 

20.  It  is  shown  in  geometry  that  the  area  of  a  triangle  is 
equal  to  half  the  proditct  of  its  base  by  its  altitude.  From 
this  theorem  find  the  base  of  that  triangle  whose  altitude 
exceeds  its  base  by  3  feet,  and  whose  area  is  35  square  feet. 


224  QUADRATIC  EqUATIONS. 

Section  III.    Equations  which  may  be  Solved 
LIKE  Quadratics. 

257.  Whenever  an  equation  contains  only  two 
powers  of  the  unknown  quantity,  and  the  index  of 
one  power  is  double  that  of  the  other,  the  equation 
can  be  solved  as  a  quadratic. 

Special  Example.     Let  us  take  tlie  e(|natioii 

2,-"  4-  M'  -f  r  =  0.  (1) 

Transposing  c  and  adding  -¥  to  each  side  of  the  equa- 

tion,,  it  becomes 

:,.«  4.  ix'  +  \¥  =  \v  -  c. 
4  4 

The  first  member  of  this  equation  is  a  i)erfect  square, 
namely,  the  square  of  a;^  -)-  —h.  Extracting  the  square  roots 
of  both  members,  we  have 

X'  +  \^'  =  /(j*'  -  '■)  =  ±  ^  V(V  -  ic). 

Hence  '.e  =  \l~h±  i/(i'  -  4c)]. 

Extracting  tlie  cube  root,  we  have 

General  Form.  We  now  generalize  this  solution  in  the 
following  way:  Suppose  we  can  reduce  an  equation  to  the 
form 

ax^''  +  hx^  +  6'  =  0, 
ill  which   the   exponent  n  may  be   any  quantity  whatever, 
entire  or  fractional. 

Mult,  by  a,  a^x^""  +  <^^^^^  ~  —  ^^• 

Compl.  square,     a^x^""  -\-  ahxJ^  +  t^^  —  j^^  ~  ^^" 


Ext.  root,  ax''  +  ~h  =  |/Q  ^'' 


jf  =  2  [-  7;  ±  |/(//-'  -  4r/^)]. 


EQUATIONS  SOLVED  LIKE  QUADRATICS.         225 

The  equation  is  now  reduced  to  the  form  of  a  binomiaL 
Extracting  the  wth  root,  we  have 


2V 


'—h±  Vb''  —  4:ac 


:)-. 


\  2a 

If  the  exponent  7i  is  a  fraction,  the  same  course  may  be 
followed. 

Suppose,  for  example, 

axi  -f-  ^2*3  +  c  =  0. 
Multiplying  by  a  and  transposing,  we  have 

Adding  —  to  both  members, 

ax^  +  aox^  4-  --  = ac. 

4       4 

The  left-hand  member  of  this  equation  is  the  square  of 

Extracting  the  square  root  of  both  members, 

,       -b±(b'-  4.ac)i 

whence  x^  = —, 

2a 

Raising  both  sides  of  this  equation  to  the  third  power  and 

extracting  the  square  root,  we  have 

b  ±{b'  -  4:ac)n^ 

2a  J  • 


,  .   b       (¥  Y      {b'-4.ac)i 


D 


EXFRCISES. 

Solve  the  equations: 

1.  x-\-  2x^  =  3.  2.  x-2x^-\-l  =  0. 

3.  x*  +  px'  +  q  =  0.  4.  x'  -  2nx'  +  ^rf  =  0, 

5.  a:"  -  4.axi  =  lc\  6.  x^  -  2xh  =  7. 

7.  4:X  +  Sxi  =  28.  8.  xi  -  4:X^  —  12. 

1        j^ 

9.   4a;i  -  lOa-^  =6.  10.  a^  -  x^""  =  1. 

11.   V^  -  2  +  Vx  =  0.  12.  'i/^+  Vi=  1. 


-226  QUADRATIC  EQUATIONS. 

258.  The  principle  of  the  preceding  method  may  be  ap- 
plied to  expressions  containing  the  unknown  quantity  when 
the  exponent  of  one  expression  is  double  that  of  the  other. 
Example.     To  solve 

{x  -  ay  -  2b{x  -a)  =  W 
We  might  solve  this  equation  in  the  usual  way,  but  an 
^easier  way  will  be  to  first  treat  x  —  a  itself  as  the  unknown 
quantity  and  solve  accordingly.    The  square  will  be  completed 
by  adding  ¥  to  each  member,  giving 

{x  -  ay  -  2b{x  -a)  +  b'  =  Sb\ 
:Ext.  root,  x-a-b=VSb  =  2  V~2b. 

x  =  a-\-{l  +  2\/2)b,      . 

EXERCISES. 

Solve  the  following  equations:* 

1.  {x  —  cy  +  2m{x  —  c)  =  3m*. 


'•  r-^)' 


C 


4.  {x"  -  cy  -  4:a{x''  -c)  =  12«'' 

5.  (x'  +  ay  +  a\x'  +  a')  =  6a' 


6.  4^__8~  =  1. 
m  in 


'.   (i-4-2g-,») 


4. 


8.  {x  -  cy  -  2(x  -  cy  =  7. 

9.  (x  -  by  -  4.b'{x  -  by  =  sb\ 

10.  x'  -  7^  -  24  +  {x'  -  7rc  +  18)^  =  0. 

259.  Process  of  Substitution.  Equations  of  the  pre- 
ceding class  may  often  be  solved  by  substituting  a  single  sym- 
l3ol  for  the  expression  containing  the  unknown  quantity. 

Example.     Solve 


e-y 


]7t(--l\  =  nn\ 


*If  the  pupil  flnds  the  management  of  these  equations  difficult,  he 
may  solve  them  by  substitution,  as  in  the  next  article. 


FACTORING   QUADRATIC  EXPRESSIONS.  227 

Let  us  put  w  = 1.  («) 

c 

The  equation  then  becomes 

ii"^  —  Snu  =  ll7i'. 
Oompl.  square,  u''  —  Snu  +  Wn'  =  ll7i^  +  16/^'  =  277^^ 

u  -  An  =  V27n  =  ±  3  i/~dn, 

u  ={4:±3  VS)n. 
Having  thus  found  the  value  of  w,  we  are  to  substitute  its. 
value  in  the  assumed  equation  (a)  in  order  to  find  the  value 
of  X.     Thus  we  have 

-  -  1  =  (4  ±  3  VS)n, 
c 

i  =z  1  +  (4  ±  3  Vd)n, 
c 

a;  =  c[l  +  (4  ±  3  V3)7i], 

EXERCISES. 

Solve  the  equations  of  the  preceding  section  by  substitu- 
tion, and  also  the  following: 


■•    \x  +  lj 


3. 


+  ly       X  + 1 

2.  (x'  -  by  +  U(x'  -  h')  =  6h\ 


Section  IV.    Factorhstg  Quadkatic  Expressions. 

360.  To  form  a  Quadratic  Equatmi  havmg  Any  Given 
Roots.  Let  us,  as  an  example,  see  how  to  form  an  equa- 
tion whose  roots  shall  be  3  and  5.  Such  an  equation  must 
be  satisfied  when  we  put 

X  =  'd     or    :^-  =  5;  (a) 

that  is,  X  —  o  =  0, 

or  .r  —  5  =  0. 

Multiplying  the  first  members,  we  have 
{x  -  3)Gt  -  5)  =  0, 
an  equation  which  is  evidently  satisfied  by  either  of  the  values 
of  X  in  {a). 


228  QUADRATIC  EqUATIONS. 

Developing  the  product,  the  equation  becomes 

x'  -  Sx  -\-  15  =  0. 
Note.    Let  the  pupil  now  solve  this  equation  and  show  that  3  and  5 
are  really  its  roots. 

EXERCISES. 

Form  equations  of  which  the  roots  shall  be: 


1.   +  3  and  -  5. 

Ans. 

:c'  +  22;  -  15  =  0. 

2.-3  and  +  5. 

3.   +  1  and  -  1. 

4.-3  and  —  8. 

5.    —  1  and  +  6. 

6.   +  3  and  +  7. 

7.  l  +  2|/3andl-2|/3. 

8.  3+  |/2and  3  - 

■   |/2. 

9.  1+  1/5  and  1  -  Vb. 

10.   5+  1/5  and  5  - 

■Vb. 

IL  1  -  |. 

12.   |and-|. 

1.1+4-1-4 

261.  Let  us  form  the  equation  of  which  the  roots  shall 
be  a  and  ^.     Proceeding  as  before, 

{x  -  a){x  -  /?)  =  0. 
Developing,  x"  —  {a  -\-  ^)  x  -\-  aft  —  0. 

Here,  the  coefficient  of  a  being  —  [a  -\-  ft),  we  see  that  a 
<|uadratic  equation  with  given  roots  is  formed  by  the  prin- 
ciple: 

The  coefficient  of  x^  is  unity. 
That  of  X  is  the  7iegative  of  the  sum  of  the  roots. 
The  term  not  containing  x  is  the  product  of  the  roots. 
The  same  is  true  of  any  quadratic  equation. 
For  we  have  found  the  roots  of  the  general  equation 
x^  -\^  jpx  ^  q  —  ^ 
to  be  —\p  -\\/{f  -  ^q) 

and  -\V-\-\  V{f  -  ^)' 

The  sum  of  these  quantities  we  find  to  be  —  p,  and  by 
multiplying  we  shall  find  their  product  to  be  q,  thus  proving 
the  proposition. 

262.  Factoring  a  Qiiadratic  Expression.  As  we  have 
Teduced  the  product 

(^  -  a){x  -  /S) 


FACTORING    qUADUATlC  EXPRESSIONS.  229 

to  the  form  x"  —  {a  -{-  I3)x  +  aj3, 

so,  conversely,  we  may  divide  the  general  expression 

:r  ^px-{-q  (b) 

into  two  factors.  In  this  expression  we  need  not  consider  x 
to  represent  an  unknown  quantity,  but  to  stand,  like  any 
other  symbol,  for  any  quantity  we  please. 

We  may  then  transform  this  expression  as  in  the  follow- 
ing examples: 

1.  Factor  x^  -{- 2x  -  d. 

We  note  that  by  §  241  the  first  two  terms,  increased  by 
unity,  will  form  a  perfect  square.  So  we  add  and  subtract 
1,  and  then  factor  by  §  238,  thus: 

x""  -\-  2x  -  S  =  x'  -}-  2x  -]-  1  -  1  -  S 
=  {x-^iy-4= 
=  {xJrl-i-2){x  +  l-2) 
=  (x-i-  3)(x  -  1). 

2.  Factor  x'  -  4:X  +  1. 
Adding  and  subtracting  4,  we  have 

X^-4:X-{-l=x'-4.X-}-4:-4:-{-l 

=  {x  -2Y  -3 

=  (x-2-i-V3){x-2-  1/3). 

EXERCISES. 


Factor: 

1.  x'  -2x-  1. 

2.  x'-{-2x-  1. 

3.  x'-^4:X-^2. 

4.  x'-]-6x-  7. 

5.  r'  -  6r  +  1. 

6.  r'  -  lOr  +  5. 

7.  p'-^Sp-  8. 

8.  r""  -%r-  2. 

363.  Let  us  now  apply  the  same  process  to  the  general 
form  (4).     We  may  transform  it  as  follows: 

x'  +  px  +  q  =  x""  +  px  +  ip'  -  ip'  +  g  1 

=  (^+ipr-{ij>'-Q)       \    (§341) 

=  {x  +  ip  +  iv{p''-ig)\{^  +  ip~ii^(p'-ii)\-   (§238) 


230  qUADMATIC  EQUATIONS. 

If  we  now  seek  the  roots  of  the  equation  formed  by  equat- 
ing these  expressions  to  zero,  we  have  the  general  principle  : 

In  order  that  a  product  may  be  equal  to  zero  at  least  one 
of  the  factors  must  be  zero. 

Hence  we  must  have  either 

whence  x  =  —  ^p  —  ^  ^{^p^  —  4g), 

or  ^-i-ip-i  V{P'  -  ^q)  =  0,  ^ 

whence  x  =  —  ip  -\-  ^  ^(p^  —  4^). 

Hence  we  may  find  the  roots  of  any  quadratic  equation 
in  the  general  form  by  factoring  the  expression  lohicJi  the 
equation  states  to  be  zero. 

EXERCISES. 

Find  the  roots  of  the  following  equations  by  factoring: 

I.  x"  -^x  =  (i. 

3.  x'  -  (a  +  b)x  =  0. 

5.  (x-iy  =  a(x'-l). 

II.  x"*  -\-  6x  -\-  S  =  0. 
9.  x""  -  X  -  I  =  0. 

11.  x'-ix-l  =  0. 

13.  x""  -  ax  -i-b  =  0. 

15.  x''-{-ax  +  b  =  0. 

17.  x^  +  mxy  -ir  f- =  0. 

264.  The  principle  of  §  261  enables  us  to  determine  the 
signs  of  the  two  roots  of  a  quadratic  equation  by  inspection. 

1.  Because  in  the  equation 

^^+pxJrq  =  0 
the  term  q  is  the  product  of  the  roots, 

q  is  positive  when  tlie  roofsi  Juire  like  signs,  and  negative 
when  they  have  unlihe  signs. 

2.  Because^  is  the  nrgalire  of  the  sum  of  the  roots, 

At  least  one  of  the  roofs  )nnst  be  negative  when  p  is  posi- 
tive, and  positive  when  p  is  negative. 
Hence  in  such  an  equation  as 

x^  +  mx  +  '?i  =  0 


2. 

x""  —  ax  =  0. 

4. 

x""  -2x-l=  0. 

6. 

x""  ^'^x-  ^  =  0. 

8. 

x'  -\-x-l  =  0. 

10. 

x'  -{-ix~2  =  0. 

12. 

x'  -ax-b  =  0. 

14. 

.t'  -{-  ax  —  b  =  0. 

16. 

x'  +  xy-f^  0. 

18. 

x'^  —  mxy  -j-  y'^  =  0. 

RELATIONS  OF  BOOTS  AND  COEFFICIENTS.        231 

the  roots,  being  like  in  signs  and  one  at  least  negative,  must 
both  be  negative. 

In  such  an  equation  as 

x^  —  mx  -|-  ^  =  0 
both  roots  must,  for  a  similar  reason,  be  positive. 
In  such  an  equation  as 

x"^  ±  7nx  —  71  =  0 
one  root  must  be  positive  and  the  other  negative. 

265.  Other  Relations  between  the  Roots  and  the  Coeffi- 
cients of  a  Quadratic  Equation.  The  preceding  theory  will 
enable  us  to  express  many  functions  of  the  roots  in  terms  of 
the  coefficients  without  solving  the  equation. 

Example  1.  Find  the  sum  of  the  squares  of  the  roots 
of  the  equation 

x'-\-bx-^'2  =  0.  {a) 

Solution.     If  we  put 

a  =  the  one  root, 
(3  =  the  other  root, 
we  have,  by  §  261, 

«  +  /?=-  5,  (h) 

^/^  =  +^.  {c) 

Squaring  the  first  of  these  equations,  doubling  the  second 
and  subtracting,  we  have 

a'  +  2a/3  -f-  y5"  .=  25 
"Zafi  =    4 

a'  -f  p'  =  21.     Ans. 

Note,     The  student  may  verify  this  result  by  solving  the  equation. 
Ex.  2.     Find  the  sum  of  the  cubes  of  the  roots  of  the 
above  equation. 

Solution.     Cubing  the  equation  (b),  we  have 
a'  +  3a' /3  +  Sa/3'  -^  /3' =  -.  125. 
The  two  middle  terms  of  the  first  member  may  be  put 
into  the  form 

da'j3  +  3aj3'  =  3a/3(a  +  /3). 
Substituting  the  values  of  the  factors  from  (b)  and  (c),  we 
have 

3a/3(a  -\.  jS)  =  -    30. 
«'-30  +  /J^=  -  125, 

a'-^r  ^"=  -    95.     Ans. 


232  QUADRATIC  EQUATIONS. 

EXERCISES. 

Find  the  values  of  the  following  functions  of  the  roots  of 
equation  {a): 

1.   a''-\-aft-{-  fi\  2.   a"  -  a^-{-  /3\ 

3.   «'  +  a'ft''  +  p.  4.   a'  -  daft  +  /?'. 

Prove  that  if  x^  and  ic,  are  the  roots  of  the  equation 
x^  -\-  px  -\-  q  =  0^ 
we  shall  have: 

5.  x^^  -\-  x^'  =  p""  —  2q. 

6.  x^^  +  x^x^-{-  x^^  =  2^'  —  q- 

7.  x^^  —  x^x^  +  x^  =  p""  —  dq. 

8.  x;  +  x^'  =  3pq  -p\ 
Find  the  values  of: 

9.  x^'  +  x;x^-\-xX'-i-x^ 
10.  a;/  —  iCjX  —  ^X'  +  ^,'- 


Section  Y.  Solutiois-  of  Irrational  Equations. 

366.  An  irrational  equation  is  one  in  which  the 
unknown  qnantity  appears  under  the  radical  sign. 

An  irrational  equation  may  be  cleared  of  fractions 
in  the  same  way  as  if  it  were  rational. 

Example.     Clear  from  fractions  the  equation 

a  h  ah  .  . 

■  —  — =  —  -.  (a) 

Vx-{-  c        Vx  —  c        Vx"  —  c^ 


The  L.C.M.  of  the  denominators  is  V{x  -\-  c){x  —  c)  = 


^x'  —  c'.    Multiplying  all  the  terms  by  this  factor,  the  equa- 
tion becomes 

a  Vx  —  c  —  b  Vx-\-c  =  db,  (5) 

and  the  equation  is  cleared  of  fractions. 

267.  To  reduce  an  irrational  equation  to  a  rational  one. 
EuLE.     1.   Clear  the  equation  of  fractions, 
2.   Transpose  terms  so  that  a  surd  expression  shall  he  a 
factor  of  one  member  of  the  equation. 


SOLUTION  OF  IBRATWNAL  EQUATIONS.  23:3 

3.  Square  both  memiers  so  as  to  rationalize  the  surd  m em- 
ber of  the  equation. 

4.  Repeat  (2)  and  (3)  until  no  surds  containing  the  un- 
known quantity  are  left. 

Example.     To  rationalize  the  equation  (Z>). 
Transposing,        a  ^x  —  c  ~h  \/x  -\-  c  -\-  ah. 
Squaring,  d\x  —  c)  =  ¥{x  -\-  c)  +  a'h''  +  2a J'  I^F+T. 

Transposing,       a^{x  —  c)  —  Jf{x  -\- c)  -\-  a^W  =  2ab^  \/x  -\-~c, 
or  ya"-  b')x  -  a'c  -  Wc  +  a^h^  =  2ab'  V^^. 

Squaring,  (a""  -  ^)V  +  2(a'b''  -  a'c  -  b'c)(a'  -  b')x 

+  {a'b'  -  a'e  -  b'cy  =  4:a'b\x  +  c), 
a  rational  equation. 

EXERCISES. 

Reduce  and  solve  the  following  equations: 

1.-4=+    '  ' 


Vx-{-4:        Vx  —  4:        S/x^  —  16 
Principal  Steps.       

9{x-4)=   40  +  «  -  12  Vx^fl, 

9aj  +  36=361  -'7Qx-^4xK 

65 
X  =  o    or    -7-. 
4 

2.  {x-{-  4)i  +  (a;  -  Sy  =  7. 

3.  (a;  +  2)*  +  2(x  +  7)*  =  8. 


4. 

-         _  ^  J-  4/^»  J,  ^. 

Vx''-\-c 

6. 

Vx-{-a  —  Vx  —  a=i  Vx. 

6. 

a  -}-  Vx""  —  ax  __  ^ 
a—  Vx""  —  ax 

7. 

\-v;^+'  +  i-o. 

1+1/^^=1 


o.   — ==z =  n. 

Vx^  +  a'  -  « 


234  qUAJJHATlC  EQUATIONS. 


9.    Vx'  +  a'  +  Vx'  -a'  =  2a. 

10.    Va'  -{-x'+  Va'  -  x'  =  a. 

11             ^               1               ^ 

_  1 

.^^-^a'-a    '     V:^-^-«^  +  «^ 

a 

12.  1/^^=^  -  Vx-'d  =  Vx. 

13.  4/(^  +  ^)  -  l^(^  +  I)  ""  ^^• 

14.  4/(4:r  +  21)  +  ^{x  +  3)  =  |/(:?:  +  8). 

15.  2  |^=  1  +  |/4M-V{?2;+T)^ 

16.  V(a  +  a;)  +  |/(a  —  a;)  =  2  Va. 

17      i/(^  +  ^)    ■  ^    yi"^  -  ^) 

Va  4-  ^{a  +  2:)         Va  -  \/{a  -  x) 

18.  \'x\  y{a-  x)  -  i/(a  +  .^•) }  =  Va  {  ^{a'  —  x^)—a\, 

19.  Va'  -  X  4-  V^'  +  X  =  a-{-i. 

20.  |/«~^^^  +  Vb  —  X  =  Va-{-b. 


21.     Va-x-{-  Vx  —  b=  Va-[-h  —  2x. 
22. 


Va  —  x^  VI 


=  |/^ 


i^a  —  a;  —  Vx  —  b  X  —  b 


23. 


Vl^-x\^  Vl-x'  _a 


Vi  +  ^-'^  +  Vi 

the  first  member 

24.    |/4«  +  Z>  -  4a:  -  2  l/^'+T^^^  =  l/^. 


Express  the  first  member  as  a  ratio  and  reduce  by  composition  and 
division. 


25.  a  Vm-^x—  b  Vm  —  x=  Vm(a''  +  J"). 

26.  V{a  +  x){x-\-b)  +  V(a  -  x)(x  -b)  =  2  Va^, 


8IMULTAJVB0UJS  QUADRATIC  EQUATIONS.         235 

Skctio]^  YI.    Simultaneous  Quadkatic  and  Other 
Equations. 

268.  Def.  Two  equations,  one  or  both  of  the  sec- 
ond degree,  between  two  unknown  quantities,  are 
called  a  pair  of  simultaneous  quadratic  equations. 

The  most  geiiei'Ml  form  of  an  equation  of  the  second  de- 
gree hetvveen  the  unknown  quantities  x  and  y  is 

ax'  -f  bxy  +  cy'  J^  dx -{- ey  +/  =  0. 

If  we  eliminate  one  unknown  quantity  between  two  such 
(.'({nations,  the  resulting  equation,  when  reduced,  will  be  of 
the  fourth  degree  with  respect  to  the  other  iinknown  quan- 
tity, and  therefore  cannot  in  general  be  solved  as  a  quadratic. 

Hut  there  are  several  cases  in  which  a  solution  of  two 
equations,  one  of  which  is  of  the  second  or  some  higher 
degree,  maybe  effected,  owing  to  some  special  relations  among 
tlie  coefficients  of  the  unknown  quantities  in  one  or  both 
equations.     The  following  are  the  most  common: 

269.  Case  I.  Whe7i  one  of  the  equations  is  of  the  first 
degree  only. 

Tin's  case  may  be  solved  thus: 

Rule.  Find  the  value  of  one  of  the  unhnoimi  quantities  in 
terms  of  the  other  from  the  equation  of  the  first  degree.  This 
ralup  heing  substituted  in  the  other  equation,  we  shall  have  a 
Huadratic  equation  from  lohich  the  other  unhnoiun  quantity 
may  he  found. 

Example.     Solve 

x-^  -  2/  =  a,  )  . 

'^x  -^   =h.\  ^""^ 

From  the  second  equation  we  find 
.  _  %  +  h^ 


(*) 


whence  x^  =  — -~^ — 

4 


Substituting  this  value  in  the  first  equation  (a)  and  re- 
if  +  Qhy  -f  9b^  =  ^a  +  2h'). 


ducing,  we  find 


236  QUADRATIC  EQUATIONS. 

Solving  this  quadratic  equation, 

y  =  -U  ±^  Va  +  %h\ 
This  value  of  y  being  substituted  in  the  equation  {h)  gives 

a;  =  -  4Z»  ±  3  Va  +  W. 

The  same  problem  may  be  solved  in  the  reverse  order  by  eliminat- 
ing y  instead  of  x.     The  second  equation  {a)  gives 

%x  -h 

If  we  substitute  this  value  of  y  in  the  first  equation,  we  shall  have  a 
quadratic  equation  in  x,  from  which  the  value  of  the  latter  quantity  can 
be  found. 

The  result  should  be  pioved  by  substituting  the  values  of  x  and  y  in 
the  first  equation. 

EXERCISES. 


Solve 

1. 

x-\-y  =  a', 

xy  =  b\ 

2. 

X  —  y  =  cr, 

X 

3. 

.x  +  ^/=:3; 

x'  -f  y'  =  29. 

4. 

X  -  y  =  3; 

x'  -  /  =  51.  • 

5. 

x-\-y  =  c; 

a^-f=f. 

6. 

X  —  y  =  2c; 

x'  -\-if  =  2b\ 

7. 

x-^  =  l; 

x'  -  3?/^  =  121. 

8. 

'"'? 

x'  -~  =  21. 

Consider  -  as  an  unknown  quantity. 

9. 

.-1  =  10; 

.^  +  i,  =  58. 

10. 

^-^  =  n; 

f  +  1  =  145. 

11. 

2/'  +  i,  =  39; 

^          X 

12. 

1        ^ 

xy  =  c;         ^ 

=  m. 

simulta:neou8  quadratic  equations.       237 

13.  ^±^'  =  1;        x'^-y'  =  o\ 

14.  ?-±^  =  ^;        ax'  +  (a  -  ^>)a;v  -  ^>2/'  =  c\ 

ax  -  hy  ^  ^._^    .^^..^ 

ex  —  dy        '  '   -^ 

16.  -^—7 —T-^-  =  -?.-;        X  —  y  =  m. 

(a-x)y  6' 

17.  x-\-  y  —  a;        t =  o  • 

'   ^         '         h  —  y  X  2 

18.  X  -\-  y  =  c'j        x'  -{-  y'  =  mxy. 

270.  Case  II.  If  each  equation  contains  only  one  term 
of  the  second  degree,  and  these  terms  are  similar,  they  may  be 
eliminated  by  the  method  of  addition  and  subtraction,  lead- 
ing to  an  equation  of  the  first  degree.  We  may  then  proceed 
as  in  Case  I. 

Example.     Solve 

2xy  —    X  —    y  —       4, 
'-hxy-\-%xArZy=  —  6. 

Multiplying  the  first  equation  by  5  and  the  second  by  3, 
and  adding,  we  have 

V)xy  —  hx  —  by  =       20 
—  lOxy  +  4a;  +  6?/  =  —  12 

~    a;+    2/=         8 
Hence  y  =  %  -\-x,  (a) 

Substituting  this  value  of  y  in  the  first  of  the  given  equa- 
tions and  reducing,  we  find 

2x'  +  14:r  -  8  =  4, 

or  x'  +    '7x         =  6. 

Solving  this  quadratic,  we  find 

x=  -  I  ±  i  V73; 
whence,  from  (a),       y  =       I  ±  i  ^'«'3. 


238  QUADRATIC  EqUATI0N8. 

EXERCISES. 

Solve: 

1.  x"  ^y  =  1',        x"  -^-'dy  -x  =  2. 

2.  a;  +  2/  +  ^'  =  22;        x  —  by  -{-  x'  =  10. 

3.  X  —  y  -{-  xy  —  a;        cc  -^  y  —  xy  =  3a. 

4.  x'  +  3xy  +  Sy  =  43;        x'  -f-  3:?-^  -  3x  =  25. 


6,  X  =  a  i^x  -\-  y;  y  ■=■!)  Vx  -\-  y. 

2i*ll\.  Case  III.  When  the  functions  of  the  unknown 
quantities  in  the  two  equations  contain  one  or  more  common 
factors,  we  may  sometimes  form  an  equation  of  lower  degree 
by  taking  the  quotient  of  the  two  equations  and  dividing  out 
the  common  factors. 

Example.        x"^  -{-xy  =  «;        y"^  +  xy  =  h. 
Factoring,  we  have        x(x-\-  y)  =  a, 
y(x  •\-y)  =  'b; 
whence,  by  taking  the  quotient, 

X      a  h 

y       V  ^      a 

The  last  equation  is  of  the  first  degree,  and  by  substitu- 
tion we  readily  find 

__      a _  h 

EXERCISES. 

Solve: 

1.  x^  +  xy"^  =  a\        y^  +  ^"^V  ~  ^' 

2.  x^y  +  xy^  =  a;        x''y  —  xy"*  =  h, 

3.  X  Vx  -\-  y  =  a^;        y  Vx  -\-  y  ==■  5'. 

4.  x-\-y  =.  m(x^  +  «/');        x  —  y  =  n{x*  +  y*), 

6.  x-\-y  =  m(x^  —  y"");        x  —  y  =.  n{x*  —  y'). 
G.  x^  +  xy""  =  a;        y*  -{-  x^y  =  5. 

7.  a;  +  ^  =  7;        x^ -\- y^  =  133. 

8.  xy  +  ;/  =  «;         rr'^  —  .^/'^  =  h. 


SIMULTANEOUS  QUADRATIC  EQUATIONS         239 

272,  Case  IV.     The  solution  may  often  be  simplified 
by  combining  the  two  equations  so  as  to  get  others  simpler  in 
form  or  admitting  of  being  factored. 
The  following  is  an  elegant  example: 
x^  -j-  y^  =  a, 
xy  =  1), 
Take  twice  the  second  equation^  and  add  and  subtract  it 
from  the  first.     Then 

x^  4-  '^^y  +  2/"  =  ^  +  ^^^     or     (^  "h  yY  =  «  +  ^h; 
X*  ~  2xy  -\-  y^  =  a  —  %h,     or     {x  —  y)-  =  a  —  2h. 

Ext.  root,  X-]-  y  =  Va  +  'Zb 

X  —  y  =  Va  —  21j 


2x  =  Va  -\-  2b  -^  Va  -  2b 
2y  =  Va-{-2b~  Va  —  2b 
Dividing  by  2,  we  have  the  values  of  x  and  y. 
As  another  example,  take  the  equations 

x^  -\-  y  =  nx, 
if  J^x=z  ny. 

If  we  take  their  difference  we  shall  find  it  divisible  by 
cc  —  y,  giving  an  equation  which  leads  to 

x-\-y  =  ri-\-l, 
or  y  =  71  -{- 1  —  X. 

Substituting  this  value  of  y  in  the  first  equation,  we  have 
a  quadratic  in  x. 

EXERCISES. 

Solve: 

1.  x^  -{-  4;?/*  =  4c*;        xy  =  ?i, 

2.  x*  —  xy  =  a^;        y"^  —  xy  =  b^, 

3.  x'  -i-  y'  -\-  X  +  y  =  62;        x^  +  y""  -  x  -  y  =  44. 

4.  x'  -\-y'  -x-Jf-y  =  49;        x'  +  y^ -{- x  -  y  =  33. 

5.  x""  -  Sxy  =  12;        xy  -  Ay'  =  8. 

6.  x'^  —  y"^  -\-  X  —  y  =  2m;        x^  —  y*  —  x  -\-  y  =  2n. 
The  following  equations  may  be  solved  by  various  com- 
binations of  the  preceding  methods: 

^    X    .   y       10  2,2       Ar. 

7.  -  -f-  -  =  ^;      ^  -\-y  =  40. 

V       X        3  '  '   -^ 


240  QUADRATIC  EQUATIONS. 

8.  x-"  +  ^'  =  74;        x^y-\-  xy  =  47. 

To  solve  this  form  a  quadratic  with  «  +  y  as  the  unknown  quantity. 

9.  x""  +  /  =  85;        x-y  -{-xy=%h, 

10.  x'-\-f-\-x-\-y  =  48;        2(x  +  ^)  =  Zxy. 

11.  x^  —  xy  -\-  y"^  =  2a';        x^  -}-  xy  -]-  y^  =  2c^. 

12.  X  -\-  y^  =  ax;        x'  -\-  y  =  ly. 

13.  x"  -{-y  Vxy  =  36;         y""  ^  x  Vxy  =  72. 

J^"  Teachers  requiring  problems  leading  to  equations  with  two  or  more  un- 
known quantities  will  find  them  in  the  Appendix. 


Section  YII.     Imaginary  Roots. 

373.  Since  the  squares  of  both  negative  and  posi- 
tive quantities  are  positive,  the  square  root  of  a  nega- 
tive quantity  cannot  be  any  quantity  which  we  have 
hitherto  considered. 

Def.  The  indicated  square  root  of  a  negative 
quantity  is  called  an  imaginary  quantity. 

The  term  imaginary  is  applied  because  in  this  case 
we  have  to  imagine  or  suppose  a  quantity  of  which 
the  square  shall  be  negative. 

Def.  The  positive  and  negative  quantities  of  al- 
gebra are  called  real. 

2*74.  Theokem.  An  imaginary  quafitity  can  always  be 
expressed  as  the  product  of  a  real  quantity  into  the  square  root 
of-l. 

Proof.  Let  —  «  be  the  negative  quantity  whose  root  is  to 
be  expressed.     Because 

—  a  =  a  X  —  1, 

we  have,  by  §  227,  

V  -  a  =  Vax  V  —  1. 

Because  a  is  positive,  Va  is  real,  so  that  the  theorem  is 
proved. 

Notation.  It  is  common  to  use  the  symbol  i  for  V  —  1, 
so  that 

i  =  V^^l. 

This  is  because  it  is  easier  to  write  i  than  V  —  \. 


IMAGIJS'AHY  ROOTS.  241 


EXERCISES. 

Express  the  square  roots: 


1.    V—  4.         Ans.  2  l/—  1     or     %L 

2.  v^To.  3.  i/rg, 


4.    1/-F.  5.    V-F. 

6.  |/- 4A;.        Alls.  2  l/^  V^     or     2kH. 

7.  l/-  9/i.  8.    V-  367? 


9.     V"-  20.        Alls.   2  .  5i  V-  1     or     2  .  5H*. 


10.  V"-  12F. 

Extract  the  square  roots  of: 

11.  —  a""  -[-  2ab  —  F.         Ans.   (a  —  b)i. 

12.  -  a'  +  4«Z>  -  4<^'.  13.    -  w'  -  2?/?/^  -  n\ 

14.    _4  +  2^-l.  15.    --!^-2^-l. 

275.  Since  the  roots  of  the  general  quadratic  equation 
x'  '-f-  ]}X  -^  q  —  ^ 

contain  the  expression  \^^f  —  4^/,  it  follows  that  if  tlie  quan- 
tity under  the  radical  sign  is  negative  there  is  no  real  root. 
In  such  cases  the  algebraic  solution  may  be  expressed  by 
imaginary  quantities. 

EXERCISES. 

Solve  the  following  quadratic  equations: 

1.  x^  _  2a:  =  -  5.         Ans.  x=\±2  V^^  =p  1  ±  2L 

2.  .r'  -4:x=  -6. 

3.  x"^  —  2ax  =  —  2«^ 

4.  x'  +  4:ax  +  Sa'  =  0. 

21 Q,  When  the  solution  of  a  problem  contains  an 
imaginary  quantity  it  shows  that  the  conditions  of  the  prob- 
lem are  impossible. 

Example.  To  find  two  numbers  of  which  the  sum  shall 
be  6  and  the  product  10. 

Solving  in  the  usual  way,  we  may  put 
X  =  one  number; 
6  —  x^  the  other. 


242  QUADRATIC  EQUATIONS. 

Then  x{Q  -  :r)  =  10, 

x'  -  Ga-  =  -  10, 

x"  —  Qx  +  9  =  -  1, 

X  —  'd  ±  V^^  ='d  ±  i. 
The  answer  being  imaginary,  there  are  no  numbers  which 
fulfil  the  conditions.  Indeed,  by  assigning  to  x  different  values 
fi'om  0  to  G  we  see  that  9  is  the  greatest  possible  product  of 
two  numbers  whose  sum  is  G.  If  one  number  is  less  than  0 
or  more  than  G  the  product  will  be  negative. 

EXERCISES. 

1.  Find  two  numbers  of  which  tlie  sum  shall  be  8  and  the 
<um  of  the  squares  n.  Then  show  that  if  n  is  less  than  32 
the  condition  is  impossible  in  real  numbers. 

Method  of  Solution.  If  x  and  S  —  x  represent  the  num- 
bers, the  conditions  of  the  problem  give 

rr  +   (8  -  xf  =  n, 
or  2:r  —  I62;  +  G4  =  n. 

Solving  this  quadratic  equation,  we  find  the  roots  to  be 


a-   =:    4    ± 


f         9 


If  71  <  e32,  the  quantity  under  the  radical  sign  is  nega- 
tive, showing  that  there  is  no  real  solution. 

2.  Find  two  numbers  of  which  the  difference  shall  be  4 
and  the  sum  of  the  squares  m,  and  show  what  is  the  least  pos- 
sible value  of  m  for  which  the  problem  is  possible. 

3.  If  the  sum  of  two  numbers  is  .s-  and  the  sum  of  their 
squares  q,  what  is  the  least  possible  value  of  </? 

4.  If  the  sum  of  two  numbers  is  10  and  their  product  p, 
what  is  the  greatest  possible  value  of  jt?? 

5.  If  the  sum  of  two  numbers  is  s,  what  is  the  greatest 
possible  value  of  their  product? 

G.  If  the  difference  of  two  numbers  is  d,  what  is  the  least 
l)ossible  value  of  the  sum  of  their  squares? 

377.  Remark.  The  so-called  imaginary  quantities  are 
unreal  only  in  the  sense  that  they  cannot  be  represented  by  the 
<jrdinary  positive  and  negative  numbers  of  algebra.  They  are, 
from  a  purely  algebraic  point  of  view,  as  real  as  other  alge- 
braic quantities,  and  are  of  tlie  greatest  use  in  the  higher 


MEMORANDA   FOR  REVIEW. 


24.^ 


Memoranda   for   Review. 

Define :  Quadratic  equation;  Pure  quadratic;  Complete 
quadratic;  Binomial  equation;  Normal  form;  General  form; 
Coefl^cients. 

Pure  Equations. 

Method  of  solving  a  binomial  equation  wlien  the  unknown 

a  power; 
quantity  appears  under  the  sign  of  \  a  root; 

a  power  of  a  root. 


Complete  Quadratic  Equations, 

That  the  equation,  if  in  the  normal  form,  may 
be  reduced  by  multiplication  so  that  the 
term  containing  the  square  of  the  unknown 
quantity  as  a  factor. shall  itself  be  a  perfect 
square. 

That  the  quadratic  expression  may  then,  by 
aggregation  in  parentheses,  be  reduced  to 
the  form  of  a  binomial,  one  of  whose  terms 
is  a  perfect  square,  while  the  other  term 
does  not  contain  the  unknown  quantity. 
Explain  -j  That  we  may  then  proceed  by  two  methods: 
I.  Transpose  the  second  term  and  ex- 
tract root. 

II.  Factor  the  binomial  and  place  eacli 
factor  equal  to  zero. 

That  by  \  }■  we  obtain  \  "^^  «'!«='*!'"'  of  the 
(II.  ( two  equations 

first  degree  in  the  unknown  quantities,  and 

thus,  by  either  method,  obtain  the  same 

pair  of  values  of  the  unknown  quantities. 

To  form  a  quadratic  equation  with  given  roots. 
Expressions  of  the  coefficients  in  terms  of  the  roots. 
Rule  for  signs  of  roots. 


244  QUADRATIC  EQUATIONS. 

Equations  wliich  may  ie  solved  as  Quadratics. 

Treatment   of   the   equation  when  it  contains   only  one 
function  of  the  unknown  quantity,  and  the  exponent  of  this 
function  in  one  term  is  double  its  exponent  in  the  other  term. 
Illustrate  by  examples,  and  give  solution: 

I.  Leaving  the  function  unchanged. 
II.   Substituting  a  symbol  for  it. 

Irrational  Equations. 
Give  rule  for  solving. 

Simultaneous  Quadratic  Equations, 

Case     I.  Form;  Eule. 

Case    II.  Form;  Rule. 

Case  III.  Form;  Rule. 

Case  IV.  Form;  Rule. 

Imaginary  Roots. 
Define  Imaginary  unit. 
Reduction  of  the  square  root  of  a  negative  quantity  to  a 


number  of  imaginary  units;  Method. 


THIED    COURSE. 


PROGEESSIOE'S    VAEIATIOISTS    KEJ) 
LOGAHITHMS. 


CHAPTER  I. 
PROGRESSIONS. 


Section  I.    Arithmetical  Progression. 

278.  Def,    An   arithmetical   progression    is    a 

series  of  terms  each  of  which  is  greater  or  less  than 
the  preceding  by  a  constant  quantity. 
Example.     The  series 

3,     5,     7,     9,     11,     13,     etc.; 

7,     4,     1,     —  2,     —  5,     etc.; 

a  -\-  I,     a,     a  —  h,     a  —  'Zb,     a  —  Sb,     etc., 
are  each  in  arithmetical  progression,  because,  in  the  first,  each 
term  is  greater  than  the  preceding  by  2;  in  the  second,  each 
is  less  than  the  preceding  by  3;  in  the  third,  each  is  less  than 
the  preceding  by  b. 

Def.  The  amount  by  which  each  term  of  an  arith- 
metical progression  is  algebraically  greater  than  the 
preceding  one  is  called  the  common  difference. 

Def.  All  the  terms  of  an  arithmetical  progression 
except  the  first  and  last  are  called  arithmetical 
means  between  the  first  and  last  as  extremes. 

Oor.  The  arithmetical  mean  of  two  quantities  is 
half  their  sum. 

Example.  The  four  numbers  5,  8,  11,  14  form  four  arith- 
metical means  between  2  and  17. 

Note.     We  put  for  brevity  A.  P.  =  arithmetical  progression. 

EXERCISES. 

1.  Write  the  first  seven  terms  of  the  progression  of  which 
the  first  term  is  8  and  the  common  difference  —  3. 

2.  Write  five  terms  of  the  prosfression  of  which  the  first 
term  is  «  —  6/i  and  the  common  difference  2;t. 


248  PBOGBESSIONS. 

3.  Write  a  progression  of  five  terms  of  which  the  middle 
term  is  x  and  the  common  difference  d. 

4.  Prove  that  if  three  terms  are  in  arithmetical  progres- 
sion the  middle  one  is  half  the  sum  of  the  extremes. 

5.  Prove  that  if  four  terms  are  in  A.  P.  the  sum  of  the  ex- 
tremes is  equal  to  the  sum  of  the  means. 

6.  Prove  that  the  sum  of  five  terms  in  A.  P.  is  five  times 
the  middle  term. 

In  the  last  three  proofs  choose  some  symbol  for  the  common  differ- 
ence, and  any  expression  you  find  most  convenient  for  the  first  term. 

Problems  in  Arithmetical  Progression. 

Let  us  put 

a,  the  first  term  of  a  progression, 

d,  the  common  difference; 

n,  the  number  of  terms; 

I J  the  last  term; 

2,  the  sum  of  all  the  terms. 
The  series  is  then 

a,     a  -\-  d,     a  -{-  2d,  .  .  .  ,  I 
Any  three  of  the  above  five  quantities  being  given,  the 
other  two  may  be  found. 

2*79.  Peoblem  I.     Given  the  first  term-,  the  common  dif- 
ference and  the  numter  of  terms,  to  find  the  last  term. 
The  first    term  is  here  a, 

second  '^      ''     ^'    a  -\-  d, 
third    ''      "     *^    ^  -I-  2d. 
The  coefficient  of  d  is,  in  each  case,  1  less  than  the  num- 
ber of  the  term.     Since  this  coefficient  increases  by  unity  for 
every  term  we  add,  it  must  remain  less  by  unity  than  the 
number  of  the  term.     Hence,  whatever  be  the  value  of  i, 

The  *th  term  \^  a  -\-  (i  —  l)d. 
Hence  when  i  =  n, 

l  =  a+  (n  -  l)d.  (1) 

From  this  equation  we  can  solve  the  further  problems: 

280.  Peoblem  II.     Given  the  last  term  ?,  the  common 
difference  d  and  the  munher  of  terms  n,  to  find  the  first  term. 


PROBLEMS  IN  ARITHMETICAL  PR00RE88I0N.     249 

The  solution  is  found  by  solving  (1)  with  respect  to  a, 
which  gives 

a  =  l-  {71-  l)d.  (2) 

281.  Problem  III.  Given  the  first  and  last  terms,  a  and 
I,  and  the  numher  of  terms  n,  to  find  the  common  difference. 

Solution  from  (1),  d  being  the  unknown  quantity, 

d  =  I^.  (3) 

282.  Problem  IV.  Given  the  first  and  last  terms  and 
the  conunon  diff^erence,  to  fi7id  the  number  of  terms. 

Solution,  also  from  (1), 

I  —  a   .   ^       I  —  a  -\-  d  ,.. 

''^^-  +  ^  =  —d—-  (*) 

283.  Generalization  of  the  Preceding  Problems.  The  solu- 
tions of  these  four  problems  all  depend  upon  the  one  general 
principle: 

The  difference  betiveen  any  two  terms  of  an  arithmetical 
progression  is  equal  to  the  common  difference  7nultiplied  by  the 
difference  of  their  numbers  in  order. 

By  the  number  in  order  of  a  term  we  mean  the  numeral 
which  expresses  its  place  in  the  series,  as  the  first,  second, 
third,  etc. 

Example.  The  difference  between  the  second  and  the 
ninth  terms  is  (9  —  2)d  =  7d. 

The  difference  between  the  first  and  the  nth.  is  {n  —  l)d, 

EXERCISES. 

In  arithmetical  progressions  there  are: 

1.  Given  common  difference  +  5,  third  term  =  14;  find 
first  term.  Ans.  First  term  =  4. 

2.  Given  fourth  term  =  b,  common  difference  =  —  3;  find 
first  seven  terms. 

3.  Given  third  term  =  a-\-Xy  fourth  term  =  «  +  2a;;  find 
first  five  terms. 

4.  Given  first  term  =  a  —  b,  sixth  term  =  6a  -{-  4^;  write 
the  six  terms. 

5.  Given  fifth  term  =  7;c  —  5^/,  seventh  term  =  9.^'  —  9?/; 
find  first  seven  terms  and  common  difference. 


250  PBOGBESSrOIfS. 

6.  If  the  first  term  is  7  and  the  common  difference  5, 
what  is  the  twelfth  term? 

7.  Given  first  term  2(i,  tenth  term  11a  —  18^;  find  com- 
mon difference. 

8.  Given  first  term  a,  fifth  term  b;  write  the  three  inter- 
vening terms. 

9.  Given  tenth  term  =  12,  twelfth  term  =  8;  find  first 
term. 

10.  Given  first  term  1,  last  term  153,  common  difference 
4;  find  the  number  of  terms. 

284.  Pkoblem  V.  To  find  the  sum  of  all  the  terms  of  an 
arithmetical  progression. 

We  have,  by  the  definition  of  ^, 

:2  =  a-\-{a^d)^{a  +  M)-^  .  .  .  .  {I-  d)^l, 
the  parentheses  being  used  only  to  distinguish  the  terms. 

Now  let  us  write  the  terms  in  reverse  order.     The  term 
before  the  last  h  I  —  d,  the  second  one  before  it  /  —  2d,  etc. 
We  therefore  have 

:2  =  l-\-{l-  d)-\-{l-U) +  («  +  6?)  +  a. 

Adding  these  two  values  of  ^  together,  term  by  term,  we 
find 

2^  =  («  +  Z)  +  («5  +  0  H-  («  +  0  + . .  •  •  +  («  +  0  +  (^  +  0. 

the  quantity  (a  +  I)  being  written  as  often  as  there  are  terms; 
that  is,  n  times.     Hence 

2^    =    7l{a    +    l)y 

Remark.     The  expression  ^^-^-,  that  is,  half  the  sum  of 

the  extreme  terms,  is  the  7nean  value  of  all  the  terms.  The 
sum  of  the  n  terms  is  therefore  the  same  as  if  each  of  them 
had  this  value. 

385.  In  the  equation  (5)  we  are  supposed  to  know  tlie 
first  and  last  terms  and  the  number  of  terms.  If  other  quan- 
tities are  taken  as  the  known  ones,  we  have  to  substitute  for 
some  one  of  the  quantities  in  (5)  its  value  expressed  by  one  of 
the  equations  (1),  (2),  (3)  or  (4).    Suppose,  for  example,  that 


PROBLEMS  IN  ARITHMETICAL,  PROGRESSION.     251 

we  have  given  only  the  last  term,  the  common  aifference  and 
the  number  of  terms;  that  is,  /,  d  and  n.  We  must  then  in 
(5)  substitute  for  a  its  value  in  (2).     This  will  give 

2  =  u  [l  -  'l^d)  =  nl  -  "i^d.  (6) 

EXERCISES. 

1.  Find  the  sum  of  the  first  100  numbers: 

1  +  2  +  3  +  4  +  ....+  100. 

2.  Find  the  sum  of  the  odd  numbers  to  99: 

1  +  3  +  5  +  7  +.....+  99. 

3.  Find  the  sum  of  the  even  numbers  to  100: 

2  +  4  +  6  +  8+....+  100. 

Note  that  the  sum  of  the  results  of  the  last  two  series  is  equal  to 
that  of  the  first. 

4.  Find  the  sum  of  the  first  n  numbers: 

1  +  2  +  3  +  4  +  .  .  .  .  +  ^^.. 

5.  Find  the  sum  of  each  of  the  progressions  of  n  numbers: 

1  +  4  +  7  +....+  3;^  -  2. 

2  +  5  +  8+..  ..+3?z-l. 

3  +  6  +  9  +  .  .  .  .  +  3ji. 

6.  If  the  fourth  term  is  14  and  the  seventh  term  26,  find 
(a)  The  sum  of  the  first  seven  terms.  [inclusive. 
(h)  The  sum  of  the  terms  from  the  fourth  to  the  eleventh 

7.  In  the  progression 

a,     a  -\-  d,     a  -\-  2df  ....«  +  (2m  +  l)d, 
find  the  separate  sums  of  the  alternate  terms 

a  +  («  +  2^)  +  (a  +  46?)  +  .  .  .  .  +  («  +  2md),  and 
«  +  c/  +  (rt  +  3d)  +  (r?  +  5^)  +  ....+[«  +  (2m  +  l)r/], 
and  take  the  difference  of  these  sums. 

8.  In  a  progression  of  five  terms  the  middle  term  is  m  and 
the  common  difference  //.  Form  the  sum  of  the  terms  and 
the  sum  of  their  squares. 

9.  The  sum  of  three  numbers  in  A.  P.  is  21  and  the  sum 
of  their  squares  197.     Find  the  numbers. 

10.  Find  five  numbers  in  A.  P.  of  which  the  sum  is  25  and 
the  sum  of  the  squares  165. 


252  PROGRESSIONS. 

Remark.  Problems  like  the  two  preceding  are  most  readily  solved 
by  taking  the  middle  term  as  the  unknown  quantity. 

11.  In  an  A.  P.  of  five  terms  the  middle  term  is  m  -and  the 
common  difference  h.  Form  the  product  of  the  first,  second., 
fourth  and  fifth  terms. 

12.  In  an  A.  P.  the  product  of  the  fourth  and  sixth  terms 
exceeds  the  product  of  the  third  and  seventh  by  48.  Find  the 
common  difference. 

13.  In  an  A.  P.  of  six  terms  the  sum  of  the  first  three 
terms  is  15  and  of  the  last  three  51.     Find  the  progression. 

14.  The  sum  of  an  A.  P.  of  six  terms  is  60,  and  the  last 
term  is  three  times  the  first.     Find  the  progression. 

15.  Find  an  A.  P.  of  seven  terms  such  that  the  sum  of  the 
last  three  terms  shall  exceed  the  sum  of  the  first  three  by  36 
and  the  product  of  the  second  and  last  shall  be  100. 

16.  By  substituting  in  the  value  (5)  of  ^  the  value  (1) 
of  If  express  the  sum  of  the  progression  in  terms  of  the  first 
term,  common  difference  and  number  of  terms. 

17.  Express  the  sum  in  terms  of  the  last  term,  common 
difference  and  number  of  terms. 

18.  Find  the  sum  of  six  terms  of  the  A.  P.  whose  first 
term  is  30  and  common  difference  —  4.  Also,  find  the  sum 
of  ten  terms  and  explain  the  equality  of  the  answers  to  the 
two  questions. 

19.  The  first  term  of  an  A.  P.  is  9,  the  common  difference 
—  1  and  the  sum  35.  Find  the  number  of  terms  and  ex- 
plain the  two  answers. 

20.  Find  seven  numbers  in  A.  P.  such  that  the  sum  of  the 
squares  of  the  first  and  seventh  shall  be  234  and  the  sum  of 
the  squares  of  the  third  and  fifth  170. 

21.  A  pedestrian  having  to  make  a  journey  of  180  miles 
goes  40  miles  the  first  day  and  diminishes  his  day's  journey  4 
miles  on  each  succeeding  day.  How  long  will  it  take  him  to 
reach  his  destination?     Explain  the  two  answers. 

22.  Find  the  sum  of  12  terms  of  the  progression  of  which 
the  first  term  is  22  and  the  common  difference  —  4. 

^^~  Teachers  requiring  additional  problems  in  arithmetical  progression  will 
find  them  in  the  Appendix. 


GEOMETRICAL  PROOBESSION.  253 


Section  II.    Geometrical  Progression. 

286.  Def.  A  geometrical  progression  consists 
of  a  series  of  terms  each  of  wMcli  is  formed  by  multi- 
plying the  term  preceding  by  a  constant  factor. 

An  arithmetical  progression  is  formed  by  continued  addi- 
tion or  subtraction;  a  geometrical  progression  by  continued 
multiplication  or  division. 

Def.  The  factor  by  which  each  term  is  multiplied 
to  form  the  next  one  is  called  the  common  ratio. 

The  common  ratio  is  analogous  to  the  common  difference 
in  an  arithmetical  progression. 

In  other  respects  the  same  definitions  apply  to  both. 

Examples. 
2,     6,     18,     54,     etc., 
is  a  progression  in  which  the  first  term  is  2  and  the  common 
ratio  3. 

^,     1.     h    h    h     etc., 
is  a  progression  in  which  the  ratio  is  |-. 

+  3,     -  6,     + 12,     -  24,     etc., 
is  a  progression  in  which  the  ratio  is  —  2. 

Note.  A  progression  Uke  the  second  one  above,  formed  by  divid- 
ing each  term  by  the  same  divisor  to  obtain  the  next  term,  is  included 
in  the  general  definition,  because  dividing  by  any  number  is  the  same 
as  multiplying  by  its  reciprocal.  Geometrical  progressions  may  there- 
fore be  divided  into  two  classes,  increasing  and  decreasing.  In  an 
increasing  progression  the  common  ratio  is  greater  than  unity  and  the 
terms  go  on  increasing;  in  a  decreasing  progression  the  ratio  is  less  than 
unity  and  the  terms  go  on  diminishing. 

Remark.  In  a  progression  in  which  the  ratio  is  negative 
the  terms  will  be  alternately  positive  and  negative. 

28 T.  Def.  A  geometrical  mean  between  two 
quantities  is  the  square  root  of  their  product. 

Def.  The  intermediate  terms  of  a  geometrical  pro- 
gression are  called  geometrical  means  between  the 
extreme  terms. 


254  PMOGBBSSIONS. 

EXERCISES. 

Form  four  terms  of  each  of  the  following  geometrical  pro- 
gressions: 

1.  First  term,  2;  common  ratio,  3. 

2.  First  term,  4;  common  ratio,  —  3. 

3.  First  term,  1;  common  ratio,  —  1. 

4.  Second  term,  f ;  common  ratio,  |. 

5.  Third  term,  |;  common  ratio,  —  J. 

6.  Third  term,  a;  common  ratio,  b. 

Problems  in  Geometrical  Progression. 

288.  In  a  geometrical  progression,  as  in  an  arithmetical 
one,  there  are  five  quantities,  any  three  of  which  determine 
the  progression  and  enable  the  other  two  to  be  found.  Tlioy 
are: 

a,  the  first  term, 
r,  the  common  ratio. 
n,  the  number  of  terms. 
If  the  last  term. 
2,  the  sum  of  the  n  terms. 
The  general  expression   for  the  geometrical  progression 
will  be 

a,     ar,     ar^,     ar^,     etc.,     to     /, 

because  each  of  these  terms  is  formed  by  multiplying  the  pre- 
ceding one  by  r. 

The  same  problems  present  themselves  in  the  two  progres- 
sions.    Those  for  the  geometrical  one  are  as  follows: 

389.  Problem  I.  Given  the  first  term,  the  common  ratio 
and  the  numher  of  terms,  to  find  the  last  term. 

The  progression  will  be 

a,     ar,     ar"^,     etc. 
We  see  that  the  exponent  of  r  is  less  by  1  than  the  number 
of  the  term,  and  since  it  increases  by  1  for  each  term  added 
it  must  remain  less  by  1,  how  many  terms  soever  we  take. 
Hence  the  ni\\  term  is 

I  =  ar^'-K  (1) 

390.  Problem  II.  Given  the  last  term,  the  co?nmon  ratio 
and  the  number  of  terms,  to  find  the  first  term. 


PMOBLKMH  IN  GEOMETlllUAL  PR0G11KSSW.\. 


,)i) 


Tlie  solution  is  found  by  dividing  both  members  of  (1)  by 
r"""^  which  gives 

«  =  7i3i-  (2) 

291.  Problem  III.     Given  the  first  term,  the  last  term 
and  the  7iumber  of  terms,  to  find  the  co7nmon  ratio. 

From  (1)  we  find  r"    ' 


.n-  1 


Extracting  the  {n  —  l)th  root  of  each  member,  we  have 


n-l 


r={^)       ■  (3) 

[The  sohition  of  Problem  IV.  requires  us  to  find  7i  from 
equation  (1),  and  belongs  to  a  higher  department  of  algebra.] 

EXERCISES. 

1.  If  the  first  term  is  a  and  the  common  ratio  p%  express 
the  second,  third,  fourth  and  ?ith.  terms. 

2.  If  the  nth  term  is  x  and  the  common  ratio  1  +  p,  ex- 
press the  first  term. 

3.  What  must  be  the  common  ratio  that  the  first  term 
may  be  x  and  the  nth  one  y? 

293.  Problem  V.  To  fitid  the  surn  of  all  n  terms  of  a 
geometrical  progression. 

We  have     2  ^=  a  -{-  ar  -\-  ar"^  -\-  etc.  -|-  <^^'"~^- 

Multiply  both  sides  of  this  equation  by  r.     We  then  have 
r'2  =  ar  +  (ir^  +  <^^^  +  etc +  ar^. 

Now  subtract  the  first  of  these  equations  from  the  second. 
It  is  evident  that  in  the  second  equation  each  term  of  the 
second  member  is  equal  to  that  term  of  the  second  member  of 
tlie  first  equation,  which  is  one  place  farther  to  the  right. 
Hence,  when  we  subtract,  all  the  terms  will  cancel  each  other 
except  tlie  first  of  the  first  equation  and  the  last  of  the  second. 

Illustration.     The  following  is  a  case  in  which  a  =  2,  r  =  3,  n  =  6 
:S  =  3  +  6  +  18  +  54  +  162  +  486. 
32  =  6  +  18  +  54  +  162  +  486  +  1458 
Subtracting,     32  -  :S  =  1458  -  2  =  1456, 
or  22  =  1456    and     2  =  728. 


956  PitoonEssioj^s. 

Returning  to  the  general  problem,  we  have  by  subtraction 

(r  -  1)2  =  ar''  -  a  =  a^r""  —  1); 

r"  —  1  1  —  r" 

whence  2  =z  a =  a .  (4) 

r  —  1  1  —  r  ^  ^ 

It  will  be  most  convenient  to  use  the  first  form  when  r  >  1 
and  the  second  when  r  <1. 

By  this  formula  we  are  enabled  to  compute  the  sum  of  the 
terms  of  a  geometrical  progression  without  actually  forming 
all  the  terms  and  adding  them. 

EXERCISES. 

1.  A  farrier  having  told  a  coachman  that  he  would  charge 
him  $3  for  shoeing  his  horse,  the  latter  objected  to  the  price. 
The  farrier  then  offered  to  take  1  cent  for  the  first  nail,  2  for 
the  second,  4  for  the  third,  and  so  on,  doubling  the  amount 
for  each  nail,  which  offer  the  coachman  accepted.  There 
were  32  nails.  Find  how  much  the  coachman  had  to  pay  for 
the  last  nail,  and  how  much  in  all. 

2.  Having  the  progression 

a  -\-  ar  -{-  ar^  -{-  ar^  +  ^^% 
what  is  the  common  ratio  of  the  new  progression  formed  by 
taking 

{a)  Every  alternate  term  of  this  progression? 

{]))  Every  mth  term? 

3.  Insert  two  geometrical  means  between  the  extremes  a 
and  «^^ 

Note.  See  Problem  TIL  and  note  that  when  two  means  are  in- 
serted we  shall  have  a  progression  of  four  terms,  and,  in  general,  that 
wlien  n  means  are  inserted  between  two  extremes  the  whole  forms  a 
G. P.  of  71 -|- 2  terms. 

4.  Insert  three  geometrical  means  between  the  extremes  m. 
and  ?/^^ 

5.  Insert  two  geometrical  means  between  the  extremes  ah 
and  a'F. 

6.  What  is  the  geometrical  mean  of  a  and  «i?  Of  a  and 
ah'^     Oia-^h  and  a  —  M 

7.  Insert  two  geometrical  means  between  ab  and  a^h^. 

8.  The  arithmetical  mean  of  two  numbers  is  6^  and  their 
geometrical  mean  is  6.     Find  the  numbers. 


PROBLEMS  IN  OEOMETRIGAL  PROQRESStON.      ^57 

9.  Find  two  numbers  whose  sum  is  30  and  the  sum  of 
whose  arithmetical  and  geometrical  means  is  24. 

10.  In  the  following  series  of  numbers  and  expressions 
state  which  are  geometrical  progressions  and  which  are  not 
such: 

(a)  1,  3,     6,     10,     etc. 

\b)  2,  4,     8,     16,     etc. 

(c)  4,  2,     i,     h  etc. 

(d)  2,  -  3,     +4,     -  6,     etc. 

(e)  a,  a%     a^,     a",     etc. 
(/')  a,  2a%     3a%     5a%     etc. 

(g)    3m\     -  6m%     12m%     -  Um'\     etc. 

11.  If  we  take  the  geometrical  progression 

a,     ar,     ar"^,     ar^,     ar*,     etc.,  (a) 

and  form  a  new  series  of  terms  by  adding  each  term  to  the 
one  next  following,  thus: 

a(l  +  r),     a(r  +  0,     a(r^  +  O.     «(^  +  O.     etc., 
is  this  new  series  or  is  it  not  a  geometrical  progression?  and 
if  it  is  such,  what  is  its  common  ratio? 

12.  If,  in  the  preceding  example,  we  form  a  new  series  by 
subtracting  each  term  of  the  progression  from  the  term  next 
following,  will  the  new  series  be  a  geometrical  progression  or 
will  it  not. 

13.  If  we  multiply  all  the  terms  of  a  G.P.  by  the  same 
constant  factor,  will  the  products  form  a  G.P. ? 

14.  If  we  add  the  same  constant  quantity  to  all  the  terms, 
will  the  sums  form  a  G.  P. 

15.  Do  the  reciprocals  of  a  G.P.  form  another  G.P.  ?  If 
so,  what  is  the  common  ratio? 

16.  What  is  the  continued  product  of  the  first  three  terms 
of  the  progression  {h),  Ex.  10?  Of  the  first  four  terms?  Of 
the  first  n  terms? 

17.  One  number  exceeds  another  by  15,  and  the  arith- 
metical mean  of  the  two  is  greater  than  the  geometrical  mean 
by  |.     What  are  the  numbers? 

18.  Find  a  progression  of  three  terms  of  which  the  first 
term  is  1  and  the  continued  product  of  the  three  terms  343. 

19.  Find  the  common  ratio  of  that  progression  of  which 


258  phogressions, 

the  sum  of  the  third  and  fourth  terms  is  one  fourth  the  sum 

of  the  first  and  second. 

20.  What  is  the  common  ratio  when  the  sum  of  the  first 

and  fourth  terms  is  to  the  sum  of  the  second  and  third  as  13 

to  5? 

Note  that  1  +  ^'^  is  divisible  by  1  +  ^• 

i^^  Teachers  requiring  additional  problems  in  geometrical  progression  will 
find  tiiem  in  the  Appendix. 

Limit  of  tlie  Sum  of  a  Progression. 

393.  Theorem.  If  the  absolute  value  of  the  common 
ratio  of  a  geometrical  pi^ogi^ession  is  less  than  U7iity,  then  there 
always  will  he  a  certain  quantity  which  the  sum  of  all  the 
terms  can  never  exceed,  no  ^natter  how  many  terms  we  take. 

Example  1.     The  sum  of  the  progression 

i  +  i  +  i  -h  etc., 

in  which  the  common  ratio  is  i,  can  never  amount  to  1,  no 
matter  how  many  terms  we  take.  To  show  this,  suppose  that 
one  person  owed  another  a  dollar  and  proceeded  to  pay  him 
a  series  of  fractions  of  a  dollar  in  geometrical  progression, 
namely, 

i.     h     h     -^-^y     etc. 

When  he  paid  him  the  i  he  would  still  owe  another  i,  when 
he  paid  the  i  he  would  still  owe  another  i,  and  so  on. 
That  is,  at  every  payment  he  would  discharge  one  half  the 
remaining  debt.  Now  there  are  two  propositions  to  be  under- 
stood in  reference  to  this  subject: 

I.   The  entire  debt  can  never  he  discharged  by  such  pay- 


For,  since  the  debt  is  halved  at  every  payment,  if  there 
was  any  payment  which  discharged  the  whole  remaining  debt, 
the  half  of  a  thing  would  be  equal  to  the  whole  of  it,  which 
is  impossible. 

II.  The  debt  can  be  reduced  belong  any  assignable  limit  by 
continuing  to  pay  half  of  it. 

For,  however  small  the  debt  may  be  made,  another  pay- 
ment will  make  it  smaller  by  one  half;  hence  there  is  no 
smallest  amount  below  which  it  cannot  be  reduced. 


LIMIT  OF  THE  8UM  OF  A   PROGBESSION.  259 

These  two  propositions,  which  seem  to  oppose  each  other,  hold  the 
truth  between  them,  as  it  were.  They  constantly  enter  into  the  higher 
mathematics,  and  should  be  w^ell  understood.  We  therefore  present 
another  illustration  of  the  same  subject. 

A  B 

I \ . \ LJ-J 

i  i  i      tV 

Ex.  2.     Suppose  AB  to  be  a  line  of  given  length.     Let  us 

go  one  half  the  distance  from  ^  to  ^  at  one  step,  one  fourth 

at  the  second,  one  eighth  at  the  third,  etc.     It  is  evident 

that  at  each  step  we  go  half  the  distance  which  remains. 

Hence  the  two  principles  just  cited  apply  to  this  case.    That  is: 

(1)  We  can  never  reach  ^  by  a  series  of  such  steps,  be- 
cause we  shall  always  have  a  distance  equal  to  the  last  step  left. 

(2)  But  we  can  come  as  near  B  as  we  please,  because  every 
step  carries  us  over  half  the  remaining  distance. 

This  result  is  often  expressed  by  saying  that  we  should  reach  B  by 
taking  an  infinite  number  of  steps.  This  is  a  convenient  form  of  ex- 
pression, and  we  may  sometimes  use  it;  but  it  is  not  logically  exact, 
because  no  conceivable  number  can  be  really  infinite.  The  assumption 
that  infinity  is  an  algebraic  quantity  often  leads  to  ambiguities  and  ditfi- 
culties  in  the  application  of  mathematics. 

294.  Def.  The  limit  of  the  sum  ^  of  a  geomet- 
rical progression  is  a  quantity  which  ^  may  approach 
so  that  its  difference  shall  be  less  than  any  quantity 
we  choose  to  assign,  but  which  ^  can  never  reach. 

Examples.     1.  The  limit  of  the  sum 

i  +  i  +  i  H-  tV  +  etc. 
is  1,  because  this  sum  fulfils  the  two  conditions 

{a)  That  it  can  never  be  as  great  as  1 ; 

{b)  That  it  can  be  brought  as  near  to  1  as  we  please  i  y 
increasing  the  number  of  terms  added. 

2.  The  point  B  in  the  preceding  figure  is  the  limit  of  all 
t\\Q  steps  that  can  be  taken  in  the  manner  described. 

The  following  i)rinciple  will  enable  us  to  find  the  limit  of 
tlie  sum  of  a  progression: 

395.  Principle.  If  r  <  1,  the  power  r""  can  be 
made  as  small  as  we  ])1ense  l)y  increasing  the  value 
of  ;?,  but  can  never  be  made  equal  to  0. 


260  PROGRESSIONS. 

Suppose,  for  instance,  that 

Then  every  time  we  multiply  by  r  we  diminish  r""  by  J  of  its 
former  value;  that  is, 

r'  =  lr   =  (1  -  i)r  =  r  -  Jr, 

r^  ~  f  r^  =  r^  —  i/-^, 

r*  =  fr'  =  r'  —  ;^r', 
etc.        etc.        etc. 

296.  Problem.  To  find  the  limit  of  the  sum  of  a  geo- 
metrical progression. 

To  solve  this  problem  we  must  take  the  expression  for  the 
sum  of  71  terms,  and  see  to  what  limit  it  approaches  when  we 
suppose  n  to  increase  indefinitely.  The  required  expression, 
as  found  in  §  292,  is 

1  —  r 
which  we  may  put  into  the  form 

1  —  r      1  —  r  ^  ' 

This  expression  for  "2,  the  sum  of  n  terms,  is  identically 

the  same  as  (4),  but  different  in  form. 

If  r  is  greater  than  unity,  the  quantity  r""  will  increase 

indefinitely  when  n  increases  indefinitely,  and  the  expression 

will  have  no  limit. 

If  r  is  less  than  unity,  then,  by  §  295,  when  n  increases 

indefinitely  r'^  will  approach  0  as  its  limit,  and  therefore  the 

expression r'^  will  also  approach  0,  so  that  we  shall  have 

Limit  of  ^  ==  --^.  (7) 

Example  1.    If  ^  =  1  and  r  =  ^,  then =  2;  and 

When  n  =  2,  l  +  i  =  2-i; 

When  n  =  ^,  1  +  |  +  |-  =  2  -  }; 

When  n  =  4.,  1  +  ^  +  ^  +  1  =  2- i; 

When  7^  =  5,  1  +  ^  +  J  +  i  -f  ^  =  2  -  -^; 
etc.  etc.  etc. 


LIMIT  OF  THE  SUM  OF  A   PROGRESSION.  261 

We  see  that  as  we  make  7i  equal  to  5,  6,  7,  etc.,  indefi- 
nitely, the  sum  of  the  series  is  less  than  2  by  ^ig-,  -^j  -^^  etc., 
indefinitely. 

Since,  by  continually  halving  a  quantity,  the  parts  can  be 
made  as  small  as  we  please,  the  limit  of  the  sum  of  the  series 
is  2. 

Ex.  2.  a  =  1;     r  =  i. 

We  now  have 


1  -  r 


and 

For  7^  =  2,  1  +  i  :==  I  -  ^; 

For  7i  =  3,  1  +  1  +  1  ^  I  _  ^3_.. 

For  7.  =  4,  1  +  i  +  i  +  ^V  =  3  _  ^^. 

etc.  etc.                        etc. 

We  see  how  as  7i  increases  the  sum  approaches  |  as  its 
limit. 

Ex.  3.  a  =  l,     r  =  -  I. 
We  then  have 


l-r       ^' 
and 

For  7i  =  2,     1-4  =  1-  3^; 

For  7^  =  3,     1  -  i  +  J  =  f  +  ^\', 
For  ^  =  4,     1  _  4  +  -1  -  ^  =  I  -  ^2_. 

For  7^  =  5,     1  -  i  +  i  -  i  +  tV  -  S  +  A; 
etc.  etc.  etc. 

We  see  that  as  n  increases  the  sums  are  alternately  greater 
and  less  than  the  limit  f ,  but  that  they  continue  to  apjiroach 
it  as  before. 

EXERCISES. 

Find  the  limits  of  the  sums  of  the  following  progressions* 

1.  T  +  7^  +  Fi'H"  ^^^-y  ^^  infinitum. 

2.  -; rTT  +  777  —  etc.,  ad  infinitum. 

4        16    '    64  -^ 


262  PB00BE88I0NS. 

^'  's^i~i  ~  (s  -  ly  +  (s-  ly  ~  ^*^'-  ^^  i^^P^^i^^^^' 

3  3^        3^ 

5.  T  +  Ti  4~  "7 3  +  6tc.,  ad  mfinitum. 

4  4'        4' 

6.  — ^2  +  "^i  ~  6tc. ,  ad  infinitum, 

0  0  0 

7.  3 — , h  71 — ^ \i.  +  71 — i Ta  +  etc.,  «a  i7inmtum. 

1  +  m    '    (1  +  m)'    '    (1  +  my    '  '  -' 

8.  - — I 7- — ■ rr  +  7- — ; r^  —  etc. ,  ttd  ififitiitum. 

1  +  m       (1  +  7ny    '    (1  +  my  -^ 

9.  From  a  cistern  of  water  half  is  drawn  into  tub  A,  half 
of  what  is  left  into  tub  B,  half  of  what  is  then  left  into  tub 
4,  and  so  on  alternately.  If  this  were  continued  without 
limit,  what  proportion  of  the  water  would  go  into  the  respec- 
tive tubs? 

10.  A  man  starts  from  a  point  A  towards  a  point  B  one 
mile  distant.  But  he  stops  at  a  point  b  two  thirds  of  the 
way  from  ^  to  ^;  then  walks  to  a  point  c  two  thirds  of  the 
way  from  b  to  A;  then  to  a  point  d  two  thirds  of  the  way 
from  c  to  i,  and  so  on  alternately,  going  in  each  direction  two 
thirds  of  the  way  back  to  the  point  from  which  he  last  set 
out.  To  what  point  would  he  continually  approach,  and  what 
is  the  limit  of  the  distance  he  could  ever  walk? 

Begin  by  drawing  a  line  to  represent  the  distance  from  A  to  B,  and 
mark  the  points  b,  c,  d,  etc.,  upon  it. 


CHAPTER  II. 

VARIATION. 

297.  Bef.  When  the  value  of  one  quantity  de- 
pends upon  that  of  another,  the  one  quantity  is  called 
a  function  of  the  other. 

Example  1.  The  time  required  for  a  train  to  perform  a 
journey  depends  upon  the  speed  of  the  train.  Hence  in  this 
case  the  time  is  a  function  of  the  spewd. 

Ex.  2.  The  value  of  a  chest  of  tea  depends  upon  its 
weight.     Hence  the  value  is  a  function  of  the  weight. 

Ex.  3.  The  weight  which  a  man  can  carry  is  a  function  of 
his  strength. 

Let  the  student,  as  an  exercise,  name  other  cases  in  which 
one  quantity  is  a  function  of  another. 

298.  When  one  quantity  -2^  is  a  function  of  a 
second  quantity  x,  then  for  every  value  of  x  there  will 
be  a  corresponding  value  of  u. 

Example  1.  If  tea  is  worth  50  cents  a  pound,  then  to  the 
quantity 

1  pound  will  correspond  the  value  50  cents; 

2  pounds  "  ''  "        "     $1.00; 

3  "       "  "  "        "     $1.50; 
etc.                   etc.  etc. 

Ex.  2.  If  a  train  has  to  make  a  journey  of  60  miles  at  a 
uniform  speed,  then  to  the  speed 

60  miles  an  hour  will  correspond  the  time  60  minutes; 

jr\       i(  a       a  a  a  a  a        oa  a 

Ar\       a  a       a  a  a  ic  a       qq  (( 

30       '*  ^*       **  **  '^  **  **     120  ** 

etc.  etc.  etc. 


264  VARIATION. 

299  Direct  Variation.  Def.  When  two  quan- 
tities are  so  related  that  any  two  values  of  the  one 
hate  the  same  ratio  as  the  corresponding  values  of  the 
other,  the  one  is  said  to  vary  directly  as  the  other. 

When  one  quantity  varies  directly  as  another,  then  by 
doubling  the  one  we  double  the  other,  by  trebling  the  one  we 
treble  the  other,  by  halving  the  one  we  halve  the  other,  etc. 

Example.  The  weight  of  an  article  varies  directly  as  the 
quantity.  When  we  halve,  double  or  treble  the  quantity,  the 
weight  will  also  be  halved,  doubled  or  trebled. 

Note.     Iu  speaking  of  direct  variation  we  may  omit  the  word  direct. 

300.  Expression  of  the  fact  that  one  quantity  y  varies 
directly  as  another  quantity  x.     Let  us  put 
a  =  any  one  value  of  y; 
c  =  the  corresponding  value  of  x; 

then,  by  definition,  we  must  have,  for  all  values  of  x-  and  y, 

X  :  a  =  y  :  c. 
This  proportion  gives 

ay  =  ex, 

whence  y  =  -x. 

^       a    . 

Here  we  may  regard  —  as  a  factor  by  which  we  multiply  any 
a 

value  of  X  to  obtain  the  corresponding  value  of  y.     Hence 
The  fact  that  one  quantity  varies  as  another  is  expressed 

by  equating  the  one  to  the  product  of  the  other  into  a  constant 

factor. 

Example.    Given  that  y  varies  as  x,  and  that 

when  X  =  3, 

then  y  =  6; 

it  is  required  to  express  y  in  terms  of  x. 

Solution.    By  definition  we  have,  for  all  values  of  x  and  y; 
y  '.  6  =  X  :  3; 
.'.     3y  =  6x; 
.-.       y  =  ix, 
which  is  the  required  expression. 


INVERSE  VARIATION.  265 

EXERCISES, 

1.  Given  that  p  varies  as  r,  and  that  when  r  =  3,  then 
jf?  =:  2;  it  is  required  to  express  j[;  in  terms  of  r,  and  to  ii»d 
the  vahies  of  p  corresponding  to  r  ==  1,  2,  5,  9,  12,  16  and 
33  respectively. 

2.  If  the  value  of  a  quantity  a  of  iron  is  represented  by  b, 
what  will  represent  the  value  of  a  quantity  x  of  iron? 

3.  Given  that  y  varies  as  x,  and  that 
when  X  —  a, 

til  en  y  ~  ^\ 

it  is  required  to  express  the  vahies  of  y  when 

(a)     X  =:  3a; 

(^)     X  =  3a  -\-  ni; 

(c)  X  =  3a  -{-  27n; 

[d)  X  =  3a  -\-  3 1)1. 

4.  Prove  that  if  y  varies  as  x,  and  that  if  we  suppose  x  to 
take  a  series  of  values  in  arithmetical  progression,  such  as 

X  :=  a, 

X  =  a  -{-  d, 

X  =  a  -}-  2d, 

X  =  a  +  3d, 
etc.      etc., 
then  the  corresponding  values  of  y  will  also  form  an  arithmet- 
ical progression. 

5.  Prove  that  if  y  varies  as  x,  the  difference  between  any 
two  values  of  y  will  have  the  same  ratio  to  the  difference 
between  the  corresponding  values  of  x  which  any  value  of  y 
has  to  the  corresponding  value  of  x. 

301.  Inverse  Variation.  One  quantity  is  said  to 
vary  inversely  as  another  when  the  ratio  of  any  two 
values  of  the  one  is  the  inverse  ratio  of  the  corre- 
sponding values  of  the  other. 

If  ?/,  corresponds  to  x^ 

and  y,  ''  ''  .r„ 

then,'  when  y  varies  inversely  as  x, 

y,  '-!/.  =  •'^.  :  ^- 
Therefore  y^x^  =  y^x^. 


266  OF  VAUIATION. 

Hence,  becftuse  x^  and  y^  as  well  as  x^  and  y^  may  be  any 
pair  of  values  of  the  quantities, 

In  inverse  variation   tlie  product  of  two  corresponding 
values  of  the  quantities  is  always  the  same. 
Hence,  also,  in  inverse  variation 

One  of  the  quantities  can  always  he  found  hy  dividing 
some  dividend  hy  the  other  quantity. 

This  dividend  is  the  product  of  any  two  corresponding 
values  of  the  quantities. 

Example.  If  ?/  =  4  when  x  —  2,  and  y  varies  inversely 
as  X,  then 

To  a;  =    2  corresponds  ?/  =  4; 
"  iz;  =    4  ''  y  =  2; 

*^  ic  =    8  ''  y  =  U 

''   2;  =  16  "  y  =  i; 

"  ic  =  32  ''         y  =  h 

etc.  etc. 

It  will  be  seen  that  each  product  of  a  value  of  x  into  the 
corresponding  value  of  y  is  8. 

EXERCISES. 

1.  If  y  varies  inversely  as  x,  and  when  ^  =  3,  ?/  =  8,  it  is 
required  to  make  a  little  table  showing  the  values  of  y  for 
X  =  1,  2,  3,  etc.,  to  12. 

2.  If  y  varies  inversely  as  x,  and  when  x  =  a,  then  y  =  h, 
it  is  required  to  express  the  values  of  y  for  a;  =  1,  :?;  =  2  and 
X  =  ^. 

3.  Given  that  u  varies  inversely  as  z,  and  that  when  5;  =  3 
the  corresponding  value  of  u  is  greater  by  5  than  it  is  when 
2;  =  4;  it  is  required  to  express  the  relation  between  u  and  z 
by  an  equation. 

Note,  We  may  solve  this  problem  with  most  elegance  by  taking 
for  the  unknown  quantity  the  product  of  any  value  of  u  by  the  corre- 
sponding value  of  z.  Let  p  represent  this  product,  and  let  «2  represent 
the  value  of  u  when  s  =  4.     We  then  have  the  two  equations 

Eliminating  u^,  we  find  p  =  60.     Hence  the  required  equation  is 

_  7)^  _60 

~  z    ~   z  ' 
uz  =  60. 


INVERSE  VARIATION.  267 

4.  Given  that  u  varies  inversely  as  z,  and  that  when 
z  =  I-  tlie  vahie  of  u  is  greater  by  1  than  it  is  when  z  =  1; 
it  is  required  to  express  the  relation  between  u  and  z. 

5.  Given  that  u  varies  inversely  as  z,  and  that  the  sum  of 
the  values  of  u  tor  z  =  1  and  2;  =  2  is  5;  it  is  required  to  ex- 
press the  relation  between  tc  and  z  by  an  algebraic  equation. 

6.  Show  that  the  quantity  of  goods  which  can  be  pur- 
chased with  a  given  sum  of  money  varies  inversely  as  the 
price. 

7.  Show  that  the  time  required  to  perform  a  journey 
varies  inversely  as  the  speed. 

302.  Def.  One  quantity  is  said  to  vary  as  the 
square  of  another  when  any  two  values  of  the  one 
have  the  same  ratio  as  the  squares  of  the  correspond- 
ing values  of  the  other. 

If  ic  and  z  and  w^  and  z^  are  pairs  of  corresponding  values, 
then  when  tc  varies  as  the  square  of  z,  we  must  always  have 

u  :  u^  =  z^  :  z^. 

In  the  same  way  as  in  direct  variation  (§  300)  may  be 
shown: 

When  one  quantity  varies  as  the  square  of  another  it  is 
equal  to  that  square  multiplied  by  some  constant  factor. 

303.  Def.  One  quantity  is  said  to  vary  inversely 
as  the  square  of  another  when  the  ratio  of  any  two 
values  of  the  one  is  the  inverse  ratio  of  the  squares  of 
the  two  corresponding  values  of  the  other. 

If  u  varies  inversely  as  the  square  of  z,  and  if  n^  is  the 
value  of  u  when  z  =  z^,  then  we  always  have 

U      \     U^      —      Z^      \     Z^y 

which  gives 


z' 
We  therefore  conclude: 

Wlien  one  quantity  varies  inversely  as  the  square  of  another 
it  is  equal  to  some  constant  quantity  divided  hy  the  square  of 
that  nf,7ip,t\ 


268  vauiatiok 

EXAMPLES     AND     EXERCISES. 

1.  It  is  found  that  the  attraction  between  two  bodies  varies 
inversely  as  the  square  of  their  distance  apart*  If  when  the 
distance  is  1  foot  the  attraction  is  represented  by  2,  it  is 
required  to  express  the  attraction  at  the  distances  2,  3,  4,  5 
and  10  feet. 

2.  Supposing  the  moon  to  be  60  radii  of  the  earth  (that 
is,  60  times  as  far  from  the  earth's  centre  as  any  point  on  the 
earth's  surface),  what  would  a  ton  of  coal  weigh  at  the  distance 
of  the  moon? 

Note.  Because  the  weight  of  the  coal  is  equal  to  the  attraction  of 
the  earth,  it  varies  inversely  as  the  square  of  its  distance  from  the 
earth's  centre. 

3.  The  area  of  a  circle  varies  as  the  square  of  its  diameter. 
If  the  diameter  of  one  circle  is  2|^  times  that  of  another,  what 
is  the  ratio  of  their  areas? 

4.  The  apparent  brilliancy  of  a  candle,  that  is,  the  amount 
of  light  which  it  will  cast  upon  a  small  surface  held  perpen- 
dicular to  the  line  from  the  candle  to  the  surface,  varies  in- 
versely as  the  square  of  the  distance.  If  we  represent  by  1 
the  brilliancy  of  the  candle  at  a  distance  of  8  feet,  what  num- 
bers will  represent  its  brilliancy  at  the  distances  of  1  inch, 
1  foot,  4  feet  and  16  feet  respectively? 

5.  The  volume  of  a  sphere  varies  as  the  cube  of  its  diam- 
eter. If  one  sphere  has  3  times  the  diameter  of  another, 
what  is  the  ratio  of  their  volumes? 

6.  If  the  diameters  of  two  spheres  are  to  each  other  as 
5:2,  what  is  the  ratio  of  their  volumes? 

304,  Combined  Variation.  The  value  of  one  quantity 
may  depend  upon  the  values  of  several  other  quantities.  The 
first  quantity  is  then  called  the  function,  and  the  others  the 
independent  variables. 

We  may  then  suppose  all  the  independent  variables  but 
one  to  be  constant,  and  the  remaining  one  to  vary,  and  ex- 
press the  law  of  variation  of  the  function. 

Having  thus  supposed  each  of  the  independent  variables 
in  succession  to  vary,  the  law  of  variation  of  the  function  is 


COMBINED  VARIATION  269 

expressed  by  stating  liow  it  varies  with  respect  to  each  inde- 
pendent variable. 

Example  1.  The  tme  required  to  perform  a  journey  is  a 
function  of  the  speed  and  of  the  dista7ice. 

If  we  suppose  the  distance  to  remain  constant  and  the 
speed  to  vary,  the  time  will  vary  inversely  as  the  speed. 

If  we  suppose  the  speed  to  remain  constant  and  the 
distance  to  vary,  the  ti7ne  will  vary  directly  as  the  distance. 

These  two  relations  are  expressed  by  saying  that  the  time 
varies  directly  as  the  distance  and  inversely  as  the  speed. 

Ex.  2.     Suppose  a  candle  at         ^  j 

C  to  illuminate  a  surface  at  /. 

If  we  suppose  the  candle  to  become  2,  3,  4  or  ?^  times  as 
bright,  the  distance  CI  remaining  constant,  the  illumination 
of  /  will  become  2,  3,  4  or  ^  times  as  great. 

If  we  suppose  that  while  the  brightness  of  the  candle 
remains  constant  the  distance  CI  becomes  2,  3  or  n  times  as 

ffreat,  the  illumination  will  be  reduced  to  — ,  —  or  — ^  of  its 

first  amounto 

These  two  facts  are  expressed  by  saying  that  the  illumina- 
tion varies  directly  as  the  brightness  of  the  candle  and  in- 
versely as  the  square  of  the  distance. 

305.  Expression  of  Comlined  Variation.  If  a  function 
u  varies  directly  as  the  quantities  j3,  q,  r,  etc.,  then,  by  §300, 
its  expression  must  contain  each  of  these  quantities  as  a  sim- 
ple factor. 

If  it  varies  inversely  as  the  quantities  u,  v,  w,  etc.,  then, 
by  §  301,  its  expression  must  contain  each  of  these  quantities 
as  a  divisor. 

Therefore  the  fact  that  u  varies  directly  as^,  q,  r  .  .  .  and 
inversely  as  u^  Vj  w  ,  .  ,  is  expressed  by  an  equation  of  the 
form 

Avar  .  .  . 
u  =  -^^ , 

uvw  .  .  . 

in  which  A  represents  some  constant  quantity,  the  value  of 
which  must  be  chosen  so  as  to  fulfil  the  conditions  of  each 
special  problem. 


270  VARIATION. 


EXERCISES. 

1.  Express  that  the  function  ^^  varies  directly  as  the  square 
of  m  and  inversely  as  the  cube  of  x.  Ans.  u  =  — ^. 

X 

2.  Express  that  the  function  s  varies  directly  as  m  and  n 
and  inversely  as  the  square  of  r, 

3.  Express  that  the  function  q  varies  directly  as  m  and 
as  the  square  of  x,  and  inversely  as  the  cube  of  h  and  as  the 
fourth  power  of  z. 

4.  If  we  represent  by  unity  the  brilliancy  with  which  a 
candle  of  brightness  unity  illuminates  a  page  4  feet  away, 
what  number  will  represent  the  illumination  of  a  page  8  feet 
away  by  a  candle  of  brightness  3  ? 

5.  If  an  electric  light  gives  as  much  light  as  2500  candles, 
at  what  distance  will  it  illuminate  the  page  of  a  book  as 
brightly  as  a  candle  5  feet  away  will  ? 

6.  At  C  is  a  candle,  and  at  E  is  E  Y  G  X 
an  electric  light  equal  to  1600  can- 
dles, distant  500  feet  from  C.  It  is  required  to  find  the 
distances  from  the  candle  to  the  two  points  X  and  Y  from 
which  the  candle  and  the  electric  light  appear  equally  bril- 
liant. 

7.  If  of  two  lights  a  feet  apart  the  brighter  gives  r  times 
as  much  light  as  the  fainter,  it  is  required  to  find  the  positions 
of  the  two  points  on  the  straight  line  joining  them  from  which 
the  two  lights  appear  equally  bright. 

8.  An  electric  light  80  feet  distant  was  found  to  throw  as 
much  light  on  a  book  as  a  candle  2  feet  distant.  If  another 
candle  twice  as  bright  is  used,  how  far  must  the  electric 
light  be  placed  to  give  as  much  light  as  this  brighter  candle 
does  at  2  feet  distance? 


CHAPTER  III. 
LOGARITH  MS. 


306.  To  every  number  corresponds  a  certain "otiier 
number  called  its  common  logarithm. 

Def.  The  common  logarithm  of  a  number  is  the 
exponent  with  which  10  must  be  affected  in  order  to 
produce  the  number. 

The  term  common  logarithm  is  used  because  there  are  other  loga- 
rithms than  the  common  ones.  Since  we  are  at  present  only  concerned 
with  the  latter,  we  shall  drop  the  adjective  common. 

To  express  the  logarithm  of  a  number  n  we  write  log  n,  so 
that 

log  n  =  the  logarithm  of  n. 
Examples. 


10° 

=  1;      .-.  log      1  =       0. 

10* 

=  10;    .'.  log    10  =       1. 

10' 

=  100;  .-.  log  100  =       2. 

10- 

■'  =  T^o;   •••   logT*o   =  -  ^. 

etc.                        etc. 

Thus  we  have 

Theoeem  I. 

The  logarithm  of  1  is  zero. 

The  logarithm  of  10  is  1. 

etc.               etc. 

(a) 


307.  If  we  call  any  number  x,  and  put 
y  =  logx, 
we  have,  by  definition, 

10^  =  X.  (b) 

If  we  suppose  y  to  increase  from  0  to  1,  a;  will  increase 
from  1  to  10.     Hence 


27^  LOOARTTHMS. 

Theorem  IL    llie  logarithm  of  a  number  hetiven  1  and  10 
is  a  positive  fraction  between  0  and  +  1. 

308.  If  we  multiply  the  equation  {h)  by  10,  we  have 

10^+1  =:  10^;; 
that  is, 

log  10a;  =  ?/  -|-  1. 
Hence 

Theorem  III.     Every  time  we  multij^ly  a  number  by  10 
we  increase  its  logarithm  by  unity. 

309.  By  dividing  (b)  by  10  we  obtain 

Theorem   IV.     Every  time  tve  divide  a  number  by  10  we 
diminish  its  logarithm  by  miity. 

310.  In  order  that  10*'  may  be  less  than  unity,  y  must  be 
negative.     Since  we  have 

10-' =  ,1, 
10 -'=.01, 
10-'=  .001, 
lO-'^^.OOOl, 
etc.         etc., 
we  see  that  as  we  increase  the  exponent  negatively  the  num- 
ber diminishes  without  limit.     Hence 

Theorem  Y.   The  logarithm  of  a  proper  fraction  is  nega- 
tive. 

Theorem  VI.   The  logarithm  of  0  is  Jiegative  infinity. 

311.  The  use  of  logarithms  is  founded  on  the  four  fol- 
lowing theorems. 

Theorem  VII.  The  logarithm  of  a  product  is  equal  to  the 
sum  of  the  logarithns  of  its  factors. 

Proof.     Let  p  and  q  be  two  factors,  and  suppose 
h  =  log  p,  h  =  log  q. 

Then  lO'^  =  p,  10^  =  q. 

Multiplying,  lO'^lO'^  =  10^^  +  ^^  =  pq. 

Whence,  by  definition, 

h  ^  h  =  log  {pq), 
or  log^  +  log  g  =  log  ( pq). 

This  proof  may  be  extended  to  any  number  of  factors. 


TBEOUEMS.  ^73 

Theorem  VIII.  The  loyarithrn  of  a  quotient  is  found  by 
subtracting  the  logaritlim  of  the  divisor  from  that  of  the  divi- 
dend. 

Proof.  Dividing  instead  of  multiplying  the  equations  in 
the  last  theorem,  we  have 

lU*^  q 

Hence,  by  definition,  k  —  k  =  log -, 

or  log^  —  log  q  =  log -. 

Theorem  IX.  The  logarithm  of  any  poiver  of  a  number 
is  equal  to  the  logarithm  of  the  number  multiplied  by  the  expo- 
nent of  the  power. 

Proof.     Let  h  =  logp,  and  let  n  be  the  exponent. 

Then  10'^  =  p. 

Eaising  both  sides  to  the  ^^th  power, 

Whence  nh  =  logjt?^ 

or  nlog  p  =  logp"^. 

Theorem  X.  The  logarithm  of  a  root  of  a  number  is 
equal  to  the  logarithm  of  the  number  divided  by  the  index  of 
the  root. 

Proof  Let  s  be  the  number,  and  let  p  be  its  nth.  root,  so 
that 

p  =  Ys         and  s  =  pl^. 

Hence  log  s  —  logjy"  =  n  \ogp.        (Th.  IX.) 

Therefore  loff  »  =  — ^, 

^  ^  n 

iogV7  =  5f-^  , 


374  LOGARITHMS. 

EXERCISES. 

Express  the  following  logarithms  in  terms  of  logp,  log  q, 
log  X,  log  y,  log  (2;  -  y)  and  log  (.t  +  y) : 

1.  \ogpy.  Ans.  logjt;  -{-  log  ^. 

2.  log  qx 

3.  log^(/?/. 

4.  \og  2j(x -{- y), 

5.  log  a;^  +  xy.      Ans.  log  ^  -|-  log  {x  -\-  y). 

6.  log  lOjo.  Ans.  logjy  +  l. 

7.  log  lOxy. 

8.  logjo^ic.  Ans.   2  log  ;j -f  log  a;. 

9.  logp^x.  10.  log^o;. 
11.  log  lOj^V.  12.  log  10;^"^^^ 

1 


13.  log  10^  14.  log 


&    10" 

15.  log-.  16.  log^,. 

q  q 

17.  log  1'.  18.  log  |.]. 

19.  log  ^?  30.  logi5i!I. 

21.  log-l^.  22.  log~^. 

*  lOji?^'  ^  100j^5' 

23.  logl/J.  24.  \ogV{x^ij). 

25.  log^*?/i  26.  logVlO. 

27.  log  Via  28.  log  Vpq. 

xy 


29.  log  i/lO(2:  -\-  y),  30.  log  ,     ^^. 

31.  log  (a:''  -  y').  32.  log  (:?:'  -  xy). 

33.  log  (2:*  -  x\f).  34.  log  jt?(:z:'  -  xhf'). 

35.  log  (a:'  -  yy.  36.  logjf?^^*(a;^  -  :ry)* 


TABLE  OF  L0OABITHM8.  276 


Table  of  Logarithms.* 

313.  The  logarithm  of  a  number  consists  of  an  integer 
and  a  decimal  fraction. 

The  decimal  fraction  is  called  the  mantissa  of  the  loga- 
rithm. 

The  integer  is  called  the  characteristic  of  the  logarithm. 

A  table  of  logarithms  gives  only  the  mantissse. 

313.  To  find  the  mantissa  of  the  logarithn  of  a  give^i 
number. 

Case  I.  When  the  number  has  three  or  fewer  figures. 

Rule.  Find  in  the  left-hand  column  that  line  which  con" 
tains  the  first  two  figures  of  the  number,  and  select  that  column 
ivhich  has  the  third  figure  at  its  top.  TJie  four  figures  i7i  this 
<iolmnn  on  the  line  selected  form  the  mantissa  of  the  logarithm. 

Examples.  Mantissa  of  log  134  =  .1271; 
*^  ''  log  17  =  .2304; 
"        "  log  707  =  .8494. 

Case  II.  When  the  number  has  four  or  more  figures. 

Rule.  Find  the  mantissa  of  the  first  three  figures  as  in 
<Jase  I. ,  and  call  this  mantissa  m.. 

Take  the  difference  D  between  this  mantissa  and  that  next 
following  it  in  the  table. 

Imagine  the  third  figure  of  the  given  numher  to  be  followed 
by  a  decimal  pointy  and  multiply  D  by  the  fourth  and  fifth 
figures  considered  as  decimals. 

Add  the  product  to  m;  the  sum  will  be  the  logarithm  re- 
quired. 

Example.     To  find  mantissa  of  log  17762. 

mant.  of  log  177  =  .2480 
Z>  =  24;  .62  X  24  =        15 


mant.  log  17762  =  .2495 


*  It  is  very  desirable  that  the  pupil  should  learn  to  use  logarithms  in 
multiplication  and  division  at  the  earliest  period  possible  in  his  studies. 
The  precepts  for  using  this  table  are  therefore  arranged  so  that  ^hey 
may  be  used  before  understanding  the  entire  theory  of  logarithms. 


276  LOGARITHMS. 

314,  Characteristics,  The  characteristic  of  a  logarithm 
is  less  by  unity  than  the  number  of  figures  preceding  the 
decimal  point  iu  the  number  to  which  it  corresponds. 

Examples.  The  mantissa  last  found  gives  the  following 
logarithms: 

log   17762  =  4.2495: 
log  1776.2  ^  3.2495: 
log  177.62  =  2.2495: 
log  1.7762  =  0.2495. 
Every  time  we  move  the  decimal  point  one  place  toward 
the  left  we  divide  the  number  by  10  and  diminish  the  loga- 
rithm by  unity,  which  leaves  the  mantissa  unchanged  (§309). 
If  the  number  is  a  decimal  fraction,  the  cliaracteristic  is 

—  1  when  the  figure  following  the  decimal  point  is  dif- 

ferent from  zero; 

—  2  when  one  ^ero  follows  the  decimal  point; 

—  3  when  two  zeros  follow  it,  and  so  on. 

In  these  cases  the  minus  sign  is  written  above  the  char- 
acteristic to  show  that  it  belongs  to  the  characteristic  alone 
and  not  to  the  mantissa. 

Examples.        log  0.17762      =  1.2495; 
log  0.017762    =2.2495; 
log  0.0017762  =  3.2495: 
etc.  etc. 

EXERCISES. 

Find  the  logarithms  of  the  following  numbers: 

1.  701.  6.   7032.  11.     1.0246. 

2.  700.  7.       70.32.  12.     2.0324. 

3.  70.  8.         0.7032.  13.  51.623. 

4.  7.  9.         0.007032.         14.   11.111. 

5.  0.7.  10.  2956.  15.     3.243. 

315.  To  find  the  number  corresponding  to  a  given  loga- 
rithm. 

Rule.  1.  Find  in  the  table,  if  possible,  the  mantissa  of 
the  given  logarithm.  The  corresponding  figures  in  the  left- 
hand  column  and  at  the  top  of  the  column  are  the  fig^ires  of 
the  required  number^ 


TABLE  OF  LOGARITHMS.  277 

2.  If  the  mantissa  is  not  found  in  the  tahle,  find  the  next 
smaller  mantissa.  The  three  correspo?iding  figures  are  the 
first  three  figures  of  the  required  number. 

3.  Take  the  excess  of  the  given  mantissa  over  the  next 
smaller  one  in  the  table  and  divide  it  by  the  difference  between 
the  tivo  consecutive  mantissce  in  the  table. 

4.  The  result  is  a  decimal  fraction  to  be  written  after  the 
three  figures  already  found,  the  decimal  'point  being  dropped. 

5.  Having  thus  found  the  figures  of  the  number,  insert  the 
decimal  point  correspondi7ig  to  the  give7i  characteristic  accord- 
ing  to  the  rule  of  §  314. 

Example  1.  Find  the  number  whose  logarithm  is  4.7267. 
We  find  in  the  tables  that  to  the  mantissa  .7267  corre- 
spond the  figures  533.    The  characteristic  being  4,  there  must 
be  five  figures  before  tlie  decimal  point,  so  that  we  have 
Number  =  53300. 
Ex.  2.  Find  the  number  whose  logarithm  is  1.2491. 
We  find  from  the  tables 

Next  smaller  mantissa,  2480;   N  =  178.     _ 
Given  mantissa,  2491 

Excess  of  given  mantissa,  11 
Difference  of  mantissas  in  table  =  2504  —  2480  =  24. 

Figures  to  be  added  to  iV^  =  ^ J  =  .  46 

Therefore  the  figures  of  the  number  are  17846.  The 
characteristic  being  1,  there  are  two  figures  before  the  deci- 
mal point,  so  that 

Number  =  17.846. 

EXERCISES. 

Find  the  numbers  corresponding  to  the  following  loga- 
rithms : 

1.  1.2032.  6.  4.0343. 

2.  0.7916.  7.  3.1922. 

3.  0.0212.  8.  2.8282. 

4.  r.0212.  9.  1.0602. 

5.  2.0212.  10.  0.9293. 


278 


TABLE  OF  FOUR- PL  ACE  LOGARITHMS. 


No. 

0  1     1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

oooo  0043 

0086 

0128  0170 

0212 

0253 

0294 

0334 

0374 

11 

12 
13 
14 
15 
16 

17 

18 
19 

0414 
0792 
II39 

146 1 
1761 
2041 

2304 

2553 
2788 

0453 

0828 

1173 
1492 
1790 
2068 

2330 

2577 
2810 

0492 
0864 
1206 

1523 
1818 
2095 

2355 
2601 

2833 

0531 
0899 
1239 

1553 
1847 
2122 

2380 
2625 
2856 

0569 

0934 
1271 

1584 
1875 
2148 

2405 
2648 

2878 

0607 
0969 
1303 
1614 
1903 
2175 
2430 
2672 
2900 

0645 
1004 
1335 
1644 

1931 
2201 

2455 
2695 

2923 

0682 
1038 
1367 

1673 

1959 
2227 

2480 
2718 
2945 

0719 
1072 
1399 
1703 

1987 
2253 

2504 
2742 
2967 

0755 
1 106 
1430 

1732 
2014 
2279 

2529 
2765 

2989 

20 

3010  !  3032 

3054 

3075 

3096 

I  3118 

1  3324 
1  3522 
3711 
1  3892 
I  4065 
!  4232 

4393 

4548 

!  4698 

3139 

3160 

318I 

3201 

21 
22 
23 
24 
25 
26 

27 

28 
29 

3222 
3424 
3617 
3802 

3979 
4150 

4314 
4472 
4624 

3243 
3444 
3636 

3820 

3997 
4166 

4330 
4487 
4639 

3263 
3464 

3655 

3838 
4014 
4183 

4346 
4502 
4654 

3284 
3483 
3674 
3856 
4031 
4200 

4362 
4518 
4669 

3304 
3502 
3692 

3874 
4048 
4216 

4378 
4533 
4683 

3345 
3541 
3729 

3909 
4082 
4249 

4409 
4564 
4713 

3365 
3560 

3747 

3927 
4099 

4265 

4425 
4579 

4728 

3385 

3579 
3766 

3945 
4116 
4281 

4440 

4594 
4742 

3404 

3598 
3784 
3962 
4133 
4298 

4456 
4609 

4757 

30 

4771 

4786 

4800 

4814 

4829 

'  4843 

4857 

4871 

4886 

4900 

31 
32 
33 
34 
35 
36 

37 

38 
39 

4914 
5051 
5185 

5315 
5441 
5563 
5682 
5798 
5911 

4928 
5065 
5198 
5328 
5453 
5575 

5694 
5809 
5922 

4942 

5079 
5211 

5340 
5465 

5587 

5705 
5821 
5933 

4955 
5092 

5224 

5353 
5478 
5599 

5717 
5832 
5944 

4969 
5105 
5237 
5366 

5490 
5611 

5729 
5843 
5955 

i  4983 
5119 
5250 

1  5378 
5502 
5623 
5740 
5855 
5966 

4997 
5132 
5263 

5391 
5514 
5635 
5752 
5866 
5977 

501 1 

5145 
5276 

5403 
5527 
5647 

5763 

5877 
5988 

5024 
5159 
5289 
5416 

5539 
5658 

5775 
5888 

5999 

5038 
5172 
5302 

5428 

5551 
5670 

5786 

5899 
6010 

40 

6021 

6031 

6042 

6053 

6064 

i  6075 

6085 

6096 

6107 

6117 

41 
42 
43 
44 
45 
46 

47 

48 
49 

6128 
6232 
6335 

6435 
6532 
6628 

6721 
6812 
6902 

6138 
6243 
6345 

6444 
6542 
6637 

6730 
6821 
691 1 

6149 
6253 
6355 

6454 
6551 
6646 

6739 
6830 
6920 

6160 
6263 
6365 

6464 
6561 
6656 

6749 
6839 
6928 

6170 
6274 
6375 
6474 

6571 
6665 

6758 
6848 
6937 

1  6180 
6284 
6385 

6484 
6580 
!  6675 
1  6767 
1  6857 
!  6946 

6191 
6294 
6395 

6493 
6590 
6684 

6776 
6866 
6955 

6201 
6304 
6405 

6503 
6599 
6693 

6785 
6875 
6964 

6212 
6314 
6415 

6513 
6609 
6702 

6794 

6884 
6972 

6222 

6325 
6425 

6522 
6618 
6712 

6803 
6893 
6981 

50 

6990 

6998 

7007 

7016 

7024 

i  7033 

7042 

7050 

7059 

7067 

51 
52 
53 
54 

7076 
7160 
7243 
7324 

7084 
7168 
7251 
7332 

7093 
7177 

7259 

7340 

7101 

7185 
7267 

7348 

7110 

7193 
7275 
7356 

1  7118 
7202 

1  7284 
'  7364 

7126 
7210 
7292 

7372 

7135 
7218 
7300 
7380 

7143 
7226 
7308 
7388 

7152 
7235 
7316 

7396 

TABLE  OF  FOUR-PLACE  LOGARITHMS 


279 


No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

55 
50 

7404 

7482 

7412 
7490 

7419 
7497 

7427 
7505 

7435 
7513 

7443 
7520 

7451 
7528 

7459 
7536 

7466 
7543 

7474 

7551 

57 

58 
59 

7559 
7634 
7709 

7566 
7642 
7716 

7574 
7649 

7723 

7582 
7657 
7731 

7589 
7664 

7738 

7597 
7672 

7745 

7604 
7679 
7752 

7612 
7686 
7760 

7619 
7694 
7767 

7627 
7701 

7774 

GO 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7S39 

7846 

01 
62 
63 

7853 
7924 

7993 

7860 

7931 

8000 

7868 
7938 
8007 

7875 
7945 
8014 

7882 
7952 
8021 

7889 

7959 
8028 

7896 
7966 
8035 

7903 
7973 
8041 

7910 
79S0 
8048 

7917 
7987 
8055 

64 
65 
66 

8062 
8129 
8195 

8069 
8136 
8202 

8075 
8142 
8209 

8082 

8149 

8215 

8089 
8156 
8222 

8096 
8162 

8228 

8102 
8169 

8235 

8109 
8176 
8241 

8116 

8182 
8248 

8122 

8189 
8254 

67 

68 
69 

8261 

8325 
8388 

8267 
8331 
8395 

8274 
8338 
8401 

8280 

8344 
8407 

8287 
8351 
8414 

8293 
8357 
8420 

8299 
8363 
8426 

8306 
8370 
8432 

8312 
8376 
8439 

8319 

8382 
8445 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8543 
8603 
8663 

8488 

8494 

8500 

8506 

71 
72 
73 

8513 
8573 
8633 

8519 
8579 
8639 

8525 
8585 
8645 

S531 
S591 
8651 

8537 
8597 
8657 

8549 
8609 
8669 

8555 
8615 

8675 

8561 
8621 
8681 

8567 
8627 
8686 

74 
75 
76 

8692 

8751 
8808 

8698 
8756 

8814 

8704 
8762 
8820 

8710 
8768 
8825 

8716 

8774 
8831 

8722 

8779 
8837 

8727 
8785 
8842 

8733 
8791 
8848 

8739 
8797. 

8854 

8745 
8802 

8859 

77 

78 
79 

8865 
8921 
8976 

8871 
8927 
8982 

8876 
8932 
8987 

8882 
8938 
8993 

8887 

8943 
8998 

8893 
8949 
9004 

8899 

8954 
9009 

8904 
8960 
9015 

8910 
8965 
9020 

8915 
8971 
9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

82 
83 

9085 
9138 
9191 

9090 

9143 
9196 

9096 

9149 
9201 

9101 

9154 
9206 

9106 

9159 
9212 

9112 
9165 
9217 

9117 
9170 
9222 

9122 

9175 
9227 

9128 
9180 
9232 

9133 
9186 
9238 

84 
85 
86 

9243 
9294 
9345 

9248 
9299 
9350 

9253 
9304 
9355 

9258 
9309 
9360 

9263 
9315 
9365 

9269 
9320 
9370 

9274 
9325 
9375 

9279 
9330 
9380 

9284 
9335 
9385 

9289 
9340 
9390 

87 
88 
89 

9395 
9445 
9494 

9400 
9450 
9499 

9405 
9455 
9504 

9410 
9460 
9509 

9415 
9465 
9513 

9420 
9469 
9518 

9425 
9474 
9523 

9430 
9479 
9528 

9435 
9484 
9533 

9440 
9489 
9538 

90 

9542 

9547 

9552 

9557 

9562  1 

9566 

9571 

9576 

9581 

9586 

91 
92 
93 

9590 
9638 
9685 

9595 
9643 
9689 

9600 
9647 
9694 

9605 
9652 
9699 

9609  ! 
9657  i 
9703 

9614 
9661 
9708 

9619 
9666 
9713 

9624 
9671 
9717 

9628 

9675 
9722 

9633 
9680 
9727 

94 
95 
96 

9731 
9777 
9823 

9736 
9782 
9827 

9741 
9786 
9832 

9745 
9791 
9836 

9750 
9795  i 
9841  ; 

9754 
9800 

9845 

9759 
9805 
9850 

9763 
9809 

9854 

9768 
9814 
9859 

9773 
9818 
9863 

97 

98 
99 

9868 
9912 
9956 

9872 
9917 
9961 

9877 
9921 

9965 

9881 
9926 
9969 

9886 
9930 
9974 

9890 
9934 
9978 

9894 
9939 
9983 

9899 
9943 
9987 

9903 
9948 
9991 

9908 

9952 
9996 

APPENDIX. 


Supplementary  Exercises. 


135.    Factoring. 

13.  2  -  3a:'  +  x\  14.  81  +  ISy'  +  y\ 

15.  x^  +  {m  -\-  n)x  +  m7i,        16.  x"  —  {m  -\-  n)x  +  wm. 
17.  ay'  -  Say  +  2a.  18.  ???a;'  -  13??za:  +  40m. 

19.  m'x'  -  Smx  +  15.  20.  b'x'  -  7b'x  +  10Z>. 

21.  m'y  +  2my  -  Sy,  22.  nY  -  StiY  -  40y. 

23.  X*  +  (2a  -  3b)x'  -  6abx\  24.  x"^^  -  4:^:"  -  12. 

26.  «V  +  8«V  +  15«V. 

28.  3y  +  ISy  +  24. 

32.   -,  +  --S. 
m       m, 

36.^^:-^. 
c  c 

{a  -  by     (a  +  by 
(a  +  by      (a  -  by 

c        c  c 

42.  16a»  +  a'2  +  §^. 


25. 

%'  +  6%^  +  Sby. 

27. 

S+»S+f- 

29. 

i.+^+«- 

31. 

2a;'  +  10^  +  8. 

33. 

c        c 

35. 

a'      ¥ 
¥      a"' 

m'r*      n's* 

37. 

n'         m'* 

39. 

a'      4«^      4«' 

x''^  x'  '^  X*' 

41. 

c-  +  c«  +  i 

282  APPENDIX. 

135^,  Factors  of  Quadrinomials,    Sometimes  an  expres- 
sion containing  four  terms  may  be  expressed  as  the  product 
of  two  binomials. 
Examples. 

ax  —  ay  -\-  bx  —  by  =  a(x  —  y)  -f  l){x  —  y) 

=    {a-^b){x-  y); 
a^  —  ar  —  at  -\-  rt  =  a{a  —  r)  —  t(a  —  r) 
=    {a  —  t)  (a  —  r). 

EXERCISES. 

Factor: 

1.  ab  -}-  ac  -{- mb  -\-  mc.  2.  a^  +  «^^^  +  «^^  +  ^^* 

3.  ex  —  cy  —  x^  -\-  xy.  4.  mV  —  m^t  —  joV  +  p^t, 

5.  a^  —  a'q  —  ap  ~\-pq.  6.  ab  —  bf'  +  a(^  —  V^ff* 

7.  ax^  +  axy  +  bxy  +  by\  8.  (a^  +  ^'):z^.y  -  ab{x^  +  y'). 

9.  (a^ -b'')pq-ab{f -q%V),  mn{x^-y^)^xy{:)i^-m^xy), 

11.  a'  +  jt?'  -  op  —  «y.  12.  A:ab  +  2^?/  -  %bx  —  xy, 

13.  6*'  +  ab{x  —  y)  —  y^xy,  14.  a^  —  a^my  -\-  a'mx  —  am^xy. 

15.  1  +  m  H h  1.  16.  2  H 1 \-mn-^l. 

17.  m^j  +  1  +  1  +  — .  18.  2  +  am  H . 

'        '  mn  am 

1  Yfl 

19.  2w  +  m'7^  +  -.  20.  m^  -\ [-  mn  + 1. 

7/1/?      mq       np       nq  mp       mq      np       nq 

23.  f  +  ^-^-*.  34.  4  +  i?  +  f  +  4. 

m    ,n      m      n  ab      a        b        ab 

MISCEI-l-ANEOUS    EXERCISES. 

Factor: 
.49  o    ^'_^' 


n' 


V      a' 


^'    4-|+^-  ^'  8«'  + 24^^  +  18:.'. 

^-    8""6-+i8-  ^-  ?"~^"^F* 

7.  a"*  ~  4a2"*  +  4a^'".  8.  3;?  +  18;^'  +  27;^'. 


FACTOBINa.  283 


9    ifl±il^  +  2^Ji-+lVlO.  i'-^. 
mn  ^wf^  n^J  9       z^ 

8  "^27*  i6      256* 

\x      al       \x      al  ao  a' 

17.  16f;_£.  18.  ^'-9!^'. 

19.  9a'^'*  +  30«&'c  +  26c\        20.  2?/''  -  4.yz  +  2;2». 
21.  4  +  8a;''  +  4:x\  22.  3z/^  +  12?/2  +  12z\ 

23.  ._L^-.-J_^.  24.^;-^-^ 


3^^ 
4r 


25. 

(x+yY       (x-yf 
3       30      75 

/VX. 

26. 

x^       a'' 

27. 

28. 

14       5 

n^       dn      9* 

29. 
31. 

{a -by      {a^hy 

w'  + 1. 

30. 
32. 

n'  -  1. 

33. 

n''  -  1. 

34. 

n^""  -  1. 

35. 

1  -  7l'\ 

36. 

m^^  -  n'\ 

37. 
39. 

«»  -  2«J  +  Z^'^  +  a  -  Z>. 

38. 
40. 

1  +  ?^^ 

41. 

-^  +  2  +  1  +  -  +  ^. 

42. 

m'  —  2?;^?^  +  Ti'*  - 

43. 

^^i 2   +    ^    +   -- 

m               m 

44. 

4          .      ^    , 

-h. 


By  what  binomial  factors  must  we  multiply  the  following 
expressions  that  the  products  may  be  binomials? 

(a)  X  —  a,  (h)  x^  +  ax  +  «^ 

\c)  re  -\-n-\-\.  (d)  m'  +  m'  +  m  +  1. 

(e)  x^  —  ax-\-  a'.  (/)  x^  —  ax^  -{-  a*:c  —  a*. 


284  APPENDIX. 

189.    Problems  in  Ratio  and  Proportion. 

38.  Find  the  ratio  of  two  numbers  whose  sum  is  to  their 
difference  in  the  ratio  m  :  n. 

39.  The  speed  of  the  steamship  Servia  is  to  that  of  the 
Bothnia  as  13  to  10;  and  the  first  steams  5  miles  farther  in  8 
hours  than  the  second  does  in  10  hours.  What  is  the  speed 
of  each? 

40.  The  speed  of  two  pedestrians  was  as  4  :  3,  and  the 
slower  was  5  hours  longer  in  going  36  miles  than  the  slower 
was  in  going  24.     What  was  the  speed  of  each? 

41.  A  chemist  had  two  vessels,  A,  containing  acid,  and  B, 
an  equal  quantity  of  water.  He  poured  one  third  the  acid 
into  the  water,  and  then  poured  one  third  of  this  mixture 
back  into  the  acid.  What  was  then  the  ratio  of  acid  to  water 
in  A? 

43.  If  24  grains  of  gold  and  400  grains  of  silver  are  each 
worth  one  dollar,  what  will  be  the  weight  of  a  coin  containing 
equal  parts  of  gold  and  silver  and  worth  a  dollar? 

43.  A  chemist  has  two  mixtures  of  alcohol  and  water,  the 
one  containing  90  per  cent,  of  alcohol,  the  other  50  per  cent. 
How  much  of  the  first  must  he  add  to  1  litre  of  the  second  to 
make  a  mixture  containing  80  j)er  cent,  of  alcohol. 

44.  It  is  a  law  of  mechanics  that  the  distances  through 
which  heavy  bodies  will  fall  in  a  vacuum  in  different  times 
are  proportional  to  the  squares  of  the  times.  If  a  body  fall  48 
feet  farther  in  2  seconds  than  in  1  second,  how  far  will  it  fall 
in  1  second?    How  far  in  t  seconds? 

45.  On  a  line  are  two  points  whose  distance  is  a.  The 
first  point  divides  the  line  into  parts  whose  ratio  is  2  :  3;  the 
second  into  parts  whose  ratio  is  5  :  7.  What  is  the  length  of 
the  line  in  terms  of  «? 

46.  If  a  line  is  divided  into  two  parts  whose  ratio  is  m  :  n^ 
what  is  the  ratio  of  the  length  of  the  whole  line  to  the  distance 
of  the  point  of  division  from  the  middle  point? 

47.  A  line  is  divided  into  three  segments  proportional  to 
the  numbers  m,  p  and  q.  What  is  the  ratio  of  the  parts  into 
which  the  middle  point  of  the  whole  line  divides  the  middle 
segment? 


RATIO  AND  PROPORTION.  285 

48.  A  sailing-ship  leaves  port,and  12  hours  later  is  fol- 
lowed by  a  steamship.  The  ratio  of  the  speeds  being  3  :  8, 
how  long  will  it  take  the  steamer  to  overtake  the  ship? 

49.  A  courier  started  from  his  post,  going  7  miles  in  3 
hours.  Two  hours  later  another  followed,  going  7  miles  in  2 
hours.     How  long  will  the  second  be  overtaking  the  first? 

50.  The  areas  of  the  openings  of  two  water-faucets  are  in 
the  ratio  3:5;  the  speeds  of  flow  of  the  water  through  the 
openings  are  in  the  ratio  3:4.  At  the  end  of  an  hour  1221 
gallons  more  have  flowed  through  the  second  than  through 
the  first.     What  was  the  flew  from  each  ? 

51.  The  speeds  of  two  trains,  A  and  B,  are  as  m  to  n,  and 
the  journeys  they  have  to  make  as  j^ :  q.  It  took  train  B  t 
hours  longer  to  make  its  journey  than  it  did  train  A.  What 
was  the  length  of  each  journey  and  the  speed  of  each  train? 

52.  A  street-railway  runs  along  a  regular  incline,  in  conse- 
quence of  which  the  speeds  of  the  cars  going  in  the  two  direc- 
tions are  as  2:3.  The  cajs  leave  each  terminus  at  regular 
intervals  of  5  minutes.  x\t  what  intervals  of  time  will  a  car 
going  up  hill  meet  the  successive  cars  coming  down,  and  vice 
versa  9 

53.  The  same  thing  being  supposed,  two  cars  starting  out 
simultaneously  from  the  two  ends  of  the  route  meet  in  30  min- 
utes.    How  long  in  time  is  the  journey  for  each  car? 

54.  Give  the  algebraic  answers  to  the  two  preceding  ques- 
tions when  the  ratio  of  the  speeds  is  di  :  n. 

55.  Find  two  numbers  which  are  to  each  other  as  4:3, 
and  whose  difference  is  J  of  the  less. 

56.  li  X  :  y'.'.Q  :%  and  ^x  —  3?/  =  7,  what  is  the  value  of 
X  and  ?/? 

57.  A  yearns  profits  were  divided  among  two  partners  in 
the  proportion  of  3  :  4.  If  the  second  should  give  $425  to  the 
first,  their  shares  would  be  equal.  What  was  the  amount 
divided? 

58.  In  a  first  yearns  partnership  A  had  3  shares,  and  B  4. 
In  the  second  A  had  1,  and  B  2.  In  the  second  year  A  had 
$30e  less  than  he  had  the  first,  and  B  had  $200  more.  What 
were  the  profits? 

50.   In  a  farm-yard  there  aic  4  slu^ep  to  every  3  cattle,  and 


286  APPENDIX. 

5  cattle  to  6  Hogs.     How  many  hogs  are  there  to  every  2i> 
sheep  ? 

60.  A  drover  started  to  market  with  a  herd  of  7  horses  to 
every  5  mules.  He  sold  27  horses  and  bought  3  mules,  and 
then  had  3  horses  to  every  4  mules.  How  many  of  each  had 
he  at  first? 

61.  Find  two  quantities  whose  sum,  difference  and  product 
are  proportional  to  5,  1  and  12  respectively. 

62.  What  number  is  that  to  which  if  2,  6  and  10  be  sever- 
ally added,  the  first  sum  shall  be  to  the  second  as  the  second 
is  to  the  third  ? 

63.  What  two  numbers  are  to  each  other  as  3  to  4,  to  each 
of  which  if  4  be  added,  the  sums  will  be  as  4  to  5? 

64.  What  quantity  must  be  taken  from  each  term  of  the 
ratio  m  :  n  that  it  may  equal  the  ratio  c  :  d^ 

05.  If  a  :  b\)Q  the  square  of  the  ratio  of  a-{-  c\  b  -\-  c,  show 
that  c  is  a  mean  proportional  between  a  and  b. 

66.  li  a  :  b  ::  c  :  d,  show  that  a{a  -\-  b  -{■  c  -{-  d)  = 
{a-^b)  {a-\-c). 

67.  A  line  is  divided  by  one  point  into  two  parts  in  the 
ratio  of  3  :  5,  and  by  another  point  into  two  parts  in  the  ratio 
of  1  :  3.  The  distance  between  the  points  of  division  is  1 
inch.     What  is  the  length  of  the  line? 

68.  One  ingot  contains  two  parts  of  gold  and  one  of  silver, 
and  another  two  parts  of  gold  and  three  of  silver.  If  equal 
parts  are  taken  from  each  ingot,  what  will  be  the  proportion 
of  the  gold  to  the  silver  in  the  alloy? 

69.  If  two  ounces  be  taken  from  the  first  of  the  above 
ingots  and  three  from  the  second,  what  will  be  the  ratio  of  the 
gold  to  the  silver? 

70.  A  cask  contains  4  gallons  of  water  and  18  gallons  of 
alcohol.  How  many  gallons  of  a  mixture  containing  2  parts 
■water  to  5  parts  alcohol  must  be  put  in  the  cask  so  that  there 
may  be  2  parts  water  to  7  alcohol? 

71.  Which  is  the  greater  ratio,  1  -^  a  :1  —  a  or  l-fa': 
1  —  a^,  a  being  positive  and  less  than  1? 

72.  Which  is  the  greater  ratio,  a^  —  ab  -]- b""  :  a'  -\-  ab  -{- b"* 
or  a^  —  a'b'^  +  ^*  '  f^''  +  f^^^"^  +  ^S  ^  and  b  having  like  signs? 

73.  What  number  must  be  taken  from  the  second  term  of 
the  ratio  2  :  34  and  added  to  the  first  that  it  may  equal  5  :  6? 


RATIO  AND  PROPORTION. 


287 


74.  It  iri  a  theorem  of  mechanics  that,  in  order  that  tv»o 
masses,  V  and  W,  at  the  ends  of  a  lever,  AB,  may  be  in  equi- 
librium, the  distances  of  their  points  of  suspension,  A  and  B, 
from  the  fulcrum,  F,  must  be  inversely  proportional  to  their 
weights;  that  is,  we  must  have 

FA. 


Weight  V 
A 


weight  W  =  FB 


Now,  if  the  length  AB  of  the  lever  is  /,  and  the  weights  of 
\  and  W  are  respectively  ni  and  n,  express  the  lengths  AF 
and  FB  of  the  arms  of  the  lever. 

75.  The  weights  at  the  ends  of  a  lever  are  8  and  13  kilo- 
grammes, and  the  fulcrum  is  3  inches  from  the  middle  of  the 
iever.     What  is  the  length  of  the  lever? 

76.  The  sum  of  the  two  weights  is  25  pounds,  and  the 
ratio  of  the  distance  of  the  fulcrum  from  the  middle  point  to 
the  length  of  the  lever  is  ;:i  :  9.     What  are  the  weights? 

77.  The  weights  are  m  and  n  (m  >  n),  and  one  arm  of  the 
lever  is  //  longer  than  the  other.    Express  the  length  of  the  lever. 

78.  A  lever  was  balanced  with  weights  of  7  and  9  kilo- 
grammes at  its  ends.  One  kilogramme  being  taken  from  the 
lesser  and  added  to  the  greater  (making  the  weights  6  and  10 
kilogrammes),  the  fulcrum  had  to  be  moved  2  inches.  What 
was  the  length  of  the  lever? 

79.  What  number  must  be  taken  from  each  term  of  the 
ratio  19  :  30  that  it  may  equal  the  ratio  1:2? 


80.   If  a  :  b::( 

:  (I,  show  that  aihw  Va'  -f  c-^  :  Vb'  +  d^ 

Find  the  ratio 

:r  :  //  from  tlio  following  proportions: 

81.       rr-f    y: 

./•    -    //  —  ///  :  H. 

82.      X  -f  2y  : 

X  —  2y  =  ?fi  :  2n. 

83.      x-^2y  : 

2x  -|-    //  =  ni  :  a. 

84.  7nx  -i-  7iy  : 

my  -\-  nx  —  p  :  q.     ' 

S5.  mx  4-  nv  : 

mx  —  ny  =    n  :  q. 

288  APPENDIX. 

272.    Quadratic  Equations  witli  Several  Un- 
known Quantities. 


14. 

^'  + !/'  =  «; 

x'  -y'  =  l). 

16. 

x^-f  =  «; 

18. 

xy  =  h, 
x''-]-y^  =  a; 

20. 

xy  =  b. 
x'  -  y'  =  m; 

X*  —  y*  =  n. 

22. 

x-\-  y 

— !-^    =  mx 

y 

15. 

ax"  +  by'    =  c; 

a'x"  +  b'y"  =  c\ 

17. 

x'-i-y'  =  152; 

xy  =  15. 

19. 

af  +  f  =  a; 

X  -\-y   =b. 

21. 

x^^f  =  M 

X  —  y  —  k. 

23. 

X  -\-  Vl^x' 
=  a' 

y  +Vl^y'  ___ 
xy  =  71.  {x  +  Vl  +  x')  (y-^  Vl  4-  y")  =  h\ 

24.  a;'  +  ?/'  +  a:  +  2/  =  ^^^;  ^5-  ^''  +  ?/'  +  ^^•  -  y  =  «  +  ^. 

^'^  —  ^^  +  :?;  —  ?/  =  ;i.        (^j''  +  y^)  (^  —  /y)  —  ^'■^^ 
26.  a;^  —  xy  =  a^y;  27.   (a:^"^  —  y^)  {'X  —  y)  —  (i\ 

xy  —  y^  =  b'^x;  {x^  +  y'')  {x  -\~  y)  —  b. 

28.  x'  -j-y'  -{-z'  =  84;         29.    x   +y    +z   =  12; 
X  -\-  y  -\-  z   =  14:;  ^y  -\-  yz  -{-  zx  =  47; 

xy  =  8.  ^^'^  4-  i/=^  -  /  =  0. 

SO.  x-\-y=    9;  ?i  +  ^=9;  31.  .-^  +  ^^  =  13;  xy  =  35; 
x'-\-u'=  52;  ?/'+v'=341„        /y  +  i;  =    9;  nv  =  18. 

32.  The  panel  in  a  door  is  12  by  18  inches,  and  it  is  to  be 
surrounded  by  a  margin  of  uniform  width  and  equal  surface 
to  the  panel.     How  wide  must  the  margin  be? 

33.  The  fore  wheel  of  a  coach  makes  6  more  revolutions 
than  the  hind  wheel  in  going  160  yards;  but  if  the  circumfer- 
ence of  each  wheel  be  increased  by  4  feet,  the  fore  wheel  will 
make  only  4  more  revolutions  in  160  yards.  What  is  the  cir- 
cumference of  each  wheel? 

34.  The  sum  of  three  numbers  is  15;  the  difference  between 
the  first  and  third  is  3  more  than  the  difference  between  the 
second  and  third,  and  the  sum  of  their  squares  is  93.  What 
are  the  numbers? 

35.  A  principal  of  ^6000  amounts  with  simple  interest  to 
$7800  after  a  certain  number  of  years.  Had  the  rate  been  1 
per  cent,  higher  and  the  time  1  year  longer,  it  would  have 
amounted  to  1720  more.     What  was  the  time  and  rate? 


QUADRATIC  EQUATIONS.  289 

36.  A  courier  left  a  town  riding  at  a  uniform  rate.  Three 
hours  afterwards  another  followed,  going  1  mile  an  hour 
faster.  Two  hours  after  the  second  another  started,  goiiig  6 
miles  an  hour.  They  arrive  at  their  destination  at  the  same 
time.     What  was  the  distance  and  rate  of  riding? 

Ans.  Dist,  =  60  or  6.     Speeds,  4,  5  and  6  or  1,  2  and  C. 

37.  In  a  right-angled  triangle  the  hypothenuse  is  5  and 
the  area  6.     What  are  the  sides? 

38.  Find  two  numbers  whose  product  is  180,  and  if  the 
greater  be  diminished  by  5  and  the  less  increased  by  3  the 
product  of  the  sum  and  difference  will  be  150. 

39.  Find  two  numbers  whose  sum  is  100  and  the  sum  of 
their  square  roots  14. 

40.  Find  two  numbers  whose  sum  is  35  and  the  sum  of 
their  cube  roots  5. 

41.  By  selling  a  horse  for  $130  I  gain  as  much  per  cent, 
as  the  horse  cost  me.     What  did  I  pay  for  him  ? 

42.  What  is  the  price  of  apples  a  dozen  when  four  less  in 
20  cents^  worth  raises  the  price  5  cents  per  dozen? 

43.  The  sum  of  the  squares  of  three  consecutive  numbers 
is  149.     What  are  the  numbers? 

44.  If  twice  the  product  of  two  consecutive  numbers  be 
divided  by  three  times  their  sum  the  quotient  will  be  f. 
What  are  the  numbers? 

45.  A  woman  bought  a  number  of  oranges  for  36  cents. 
If  she  had  bought  2  more  for  the  same  money  she  would  have 
paid  \  of  a  cent  less  for  each  orange.     How  many  did  she  buy? 

46.  In  mowing  60  acres  of  grass,  5  days  less  would  have 
been  sufficient  if  2  acres  more  a  day  had  been  mown.  How 
many  acres  were  mown  per  day? 

47.  A  broker  bought  a  certain  number  of  shares  (par 
value  $100  each)  at  a  discount  for  $6400.  When  they  were 
at  the  same  per  cent,  premium,  he  sold  all  but  20  for  $7200. 
How  many  shares  did  he  buy,  and  at  what  price  ? 

48.  If  the  length  and  breadth  of  a  rectangle  were  each 
increased  by  2,  the  area  would  be  238;  if  both  were  each  di- 
minished by  2,  the  area  would  be  130.  Find  the  length  and 
breadth. 

49.  Twice  the  product  of  two  digits  is  equal  to  the  number 
itself;  and  7  times  the  sum  of  the  digits  is  equal  to  the  number 


290  APPENDIX. 

50.  The  sum  of  two  numbers  is  |  of  the  greater,  and  the 
difference  of  their  squares  is  45.     What  are  the  numbers? 

51.  The  numerator  and  denominator  of  two  fractions  are 
each  greater  by  2  than  those  of  another;  and  the  sum  of  the 
two  fractions  is  2f ;  if  the  denominators  were  interchanged, 
the  sum  of  the  two  fractions  would  be  3.  What  are  the  frac- 
tions? 

52.  A  man  starts  from  A  to  go  to  B.  During  the  first 
half  of  the  journey  he  drives  ^  mile  an  hour  slower  than  the 
other  half,  and  arrives  in  5|  hours.  On  his  return  he  travels 
a  mile  slower  during  the  first  half  than  when  he  went  in  go- 
ing over  tlie  same  portion,  and  returned  in  6f  hours.  What 
was  the  distance  and  rate  of  driving? 

53.  A  person  who  has  $8800  invests  a  part  of  it  in  one 
enterprise  and  the  rest  in  another;  the  dividends  differ  in 
rate,  but  are  equal  in  amount.  If  the  sums  invested  had  ex- 
changed rates  of  dividends,  the  first  would  have  yielded  $200 
and  the  other  $288.     What  were  the  rates? 

54.  Divide  50  into  two  such  parts  that  their  product  may 
be  to  the  sum  of  their  squares  as  6  to  13. 

55.  A  company  at  a  hotel  had  $12  to  pay,  but  before  set- 
tling 2  left,  when  those  remaining  had  30  cents  apiece  more 
to  pay  than  before.     How  many  were  there? 

56.  A  and  B  set  out  from  two  towns  which  are  144  miles 
apart,  and  travelled  until  they  met.  A  went  8  miles  an  hour, 
and  the  number  of  hours  they  travelled  was  three  times  greater 
than  the  number  of  miles  B  travelled  an  hour.  At  what  time 
did  they  meet,  and  what  was  B's  speed? 

57.  In  a  purse  containing  28  pieces  of  silver  and  nickel, 
each  silver  coin  is  worth  as  many  cents  as  there  are  uickel 
coins,  and  each  nickel  is  worth  as  many  cents  as  there  are 
silver  coins,  and  the  whole  are  worth  $1.50.  How  many  are 
there  of  each? 

58.  Find  two  such  numbers  that  the  product  of  their  sum 
and  difference  may  be  7,  and  the  product  of  the  sum  and  dif- 
ference of  their  squares  may  be  144. 

59.  A  grocer  sold  50  pounds  of  pepper  and  80  pounds  of 
ginger  for  $26;  but  he  sold  25  pounds  more  of  pepper  for  $10 
than  he  did  of  ginger  for  $4.  What  was  tlie  price  ]>ei"  pound 
of  each? 


ARITHMETICAL  PR00BE8SI0N.  291 


2S5,    Aritliiiietical  Progression. 

23.  Find  the  ni\i  term  of  the  series  1,  3,  6,  7,  etc. 

24.  Find  the  sum  of  n  terms  of  the  series  2,  4,  6,  8,  etc. 

25.  Find  the  sum  of  n  terms  of  the  series  1,  3,  5,  7,  etc. 
2t).  Insert  six  arithmetical  means  between  'I  and  2o. 

27.  If  a  body  fall  16  feet  during  the  first  second,  and  32 
feet  each  second  thereafter  more  than  in  the  immediately 
preceding  second,  how  far  will  it  fall  during  the  tenth  second, 
and  how  far  in  ten  seconds? 

28.  Find  the  sum  of  n  terms  of  the  series  5,  12,  19,  26, 
33,  etc. 

13 

29.  Find  the  sum  of  12  terms  of  the  series  7,  -x-,  6,  etc. 

30.  What  is  the  expression  for  the  sum  of  n  terms  of  a 

3 
series  whose  first  term  is  —  and  the  difference  between  the 

2 

third  and  seventh  terms  3? 

31.  The  difference  between  the  first  and  tenth  term  of  an 
increasing  A.  P.  is  18,  and  the  sum  of  the  ten  terms  is  100. 
What  is  the  A.  P.  ? 

32.  a  =  3,  and  the  fourth  term  is  4,  What  is  the  sum  of 
eight  terms? 

33.  There  are  two  arithmetical  series  which  have  the  same, 
common  difference;  the  first  terms  are  2  and  3  respectively, 
and  the  sum  of  five  terms  of  one  is  to  the  sum  of  five  terms  of 
the  other  as  8  :  9.     What  are  the  series? 

34.  There  are  four  numbers  in  A.  P.  whose  sum  is  to  the 
sum  of  their  squares  as  2  :  12,  and  the  sum  of  the  first  three 
is  12.     What  is  the  A.  P.  ? 

35.  A  number  consisting  of  three  digits,  which  are  in 
arithmetical  progression,  if  divided  by  the  sum  of  its  digits 
gives  15  for  a  quotient,  and  if  396  be  added  to  it  the  digits 
will  be  reversed.     What  is  the  number? 

36.  Find  three  numbers  in  arithmetical  progression  the 
isumof  whose  squares  shall  be  261,  and  the  square  of  the  mean 
greater  than  the  product  of  the  extremes  by  9. 

37.  Find  four  numbers  in  arithmetical  progression  whose 
fium  is  16  and  continued  product  105. 


•292  APPENDIX. 

38.  A  starts  from  a  certain  place  going  2  miles  the  first 
day,  tl  the  second,  G  the  third,  and  so  on;  three  days  after  B 
sets  out  and  travels  15  miles  a  day.  How  many  days  be- 
fore B  overtakes  A? 

39.  A  traveller  started  from  a  certain  place  going  2  miles 
tlie  first  day,  5  the  second,  and  so  on.  After  two  days  another 
followed  and  went  6  miles  the  first  day,  10  the  second,  and  so 
on.     After  how  many  days  will  they  be  together? 

40.  In  a  series  of  five  terms  the  sum  of  the  first  three  is 
24,  and  the  last  three  42.     What  is  the  series? 

41.  Find  three  numbers  in  arithmetical  progression  whose 
sum  shall  be  36,  and  the  sum  of  the  first  and  second  shall  be 
4  of  the  sum  of  the  second  and  third. 

42.  How  many  means  must  be  inserted  between  5  and  17 
in  order  that  the  sum  of  the  first  two  shall  be  to  the  sum  of 
the  second  two  as  4  :  7? 

43.  In  an  arithmetical  progression  of  three  terms  whose 
common  difference  is  4,  the  product  of  the  second  and  third  is 
greater  by  12  than  the  product  of  the  first  and  second.  What 
is  the  series? 

44.  The  gum  of  the  squares  of  the  extremes  of  an  arithmet- 
ical progression  of  four  terms  is  153,  and  the  sum  of  the  squares 
of  the  means  117.     What  is  the  series? 

45.  The  sum  of  five  terms  of  an  arithmetical  progression  is 
50,  and  the  product  of  the  first  and  fifth  is  to  the  product  of 
the  second  and  fourth  as  7  :  8.     What  is  the  series? 

46.  The  sum  of  nine  terms  of  an  arithmetical  progression  is 
1)0.     What  is  the  middle  term,  and  the  sum  of  the  extremes? 

47.  A  and  B  start  together  from  the  same  place;  A  goes 
20  miles  a  day  and  B  15  miles  the  first  day  and  |  mile  more 
on  each  succeeding  day  than  on  the  preceding  day.  How  far 
apart  will  they  be  at  the  end  of  10  days,  and  which  will  be  in 
advance? 

48.  A  and  B  have  a  distance  of  27  miles  to  walk;  A  starts 
at  2^  miles  an  hour  with  an  hourly  increase  of  i  mile;  B  starts 
at  5  miles  an  hour,  but  falls  off  i  mile  every  hour.  Which 
will  finish  first,  and  by  how  much? 

49.  Find  a  progression  of  four  terms  in  which  the  sum  of 
the  extremes  is  21  and  the  product  of  the  means  104. 


GEOMETIUCAL  PROGUESSIOJS^.  ^9;J 

39^.     Geometrical    Progression. 

21-  Find  three  numbers  in  A.  P.  which  being  increased 
by  1,  3,  and  10  respectively  the  sum  will  form  a  geometrical 
progression  whose  common  ratio  is  2. 

22.  Find  three  numbers  in  geometrical  progression  wliose 

19 
sum  is  19  and  the  sum  of  whose  reciprocals  --. 

oo 

23.  Express  the  sum  of  n  terms  of  the  G.  P.  whose  first 
>ierm  is  a  and  common  ratio  W, 

24.  Express  the  sum  of  n  terms  when  the  second  term  is 
2  and  the  third  is  —  6. 

25.  Show  that  the  sum  of  ^n  terms  when  divided  by  the 
«um  of  n  terms  gives  the  quotient  r'^  -\-l. 

26.  Find  five  terms  in  a  geometrical  progression  the  sum 
*f  whose  three  means  is  156  and  the  sum  of  the  two  extremes 
328. 

27.  In  a  geometrical  progression  of  three  terms,  the  sum 
of  the  first  and  second  exceeds  the  third  by  2,  and  the  sum  of 
\he  first  and  third  exceeds  the  second  by  14,  What  is  the 
series? 

28.  The  sum  of  two  numbers  is  30,  and  the  arithmetical 
mean  exceeds  the  geometrical  by  3.     What  are  the  numbers? 

29.  Having  a  progression  of  an  odd  number  of  terms, 
show  that  if  from  the  sura  of  the  even  terms  (the  second, 
fourth,  etc.)  we  subtract  the  sum  of  all  the  odd  terms  except 
the  last  (the  first,  third,  etc.),  the  remainder  is  equal  to  the 
difference  between  the  first  and  last  terms  divided  by  r  -f-  1- 

30.  In  a  geometrical  progression  of  four  terms  the  first  is 
less  than  the  third  by  24,  and  the  sum  of  the  extremes  is  to 
the  sum  of  the  means  as  7  :  3.     What  is  the  series? 

31.  The  sum  of  three  numbers  in  geometrical  progression 
is  38,  and  the  product  of  the  mean  by  the  sum  of  extremes  is 
312.     What  is  the  series? 

32.  Find  a  geometrical  progression  of  three  terms  whose 
product  is  1728  and  the  difference  of  extremes  45. 

33.  The  sum  of  four  terms  of  a  geometrical  progression  is 
80,  and  the  last  term  divided  by  the  sum  of  the  two  means  is 
2 J.     What  is  the  series? 


294  APPKMJIX. 

34.  If  1,  3  and  7  be  added  to  three  consecutive  terms  of 
an  arithmetical  progression  whose  0.  R.  is  2,  the  sums  will 
form  a  geometrical  progression.     Find  the  series? 

35.  Find  four  numbers  in  a  geometrical  progression  such 
that  the  fourth  shall  be  144  more  than  the  second,  and  the 
sum  of  the  means  be  to  the  sum  of  the  extremes  as  3  :  7. 

36.  Find  a  geometrical  progression  in  which  the  sum  of  the 
second  and  third  terms  is  60,  and  of  the  first  and  third  50. 

37.  The  sum  of  the  first  and  second  of  four  terms  of  a 
geometrical  progression  is  16,  and  the  sum  of  the  third  and 
fourth  terms  144.     What  is  the  series? 

38.  Show  that  the  difference  of  any  two  terms  of  a  Gr.  P. 
is  divisible  by  r  -  1.     (Of.  §  136.) 

39.  Find  a  geometrical  progression  of  three  terms  whose 
product  is  729,  and  the  sum  of  the  extremes  divided  by  the 
means  is  3^. 

40.  The  difference  of  two  numbers  is  48,  and  the  arith- 
metic mean  exceeds  the  geometric  by  18.  What  are  the 
numbers? 

41.  The  fifth  term  of  a  G:  P.  exceeds  the  first  by  16,  and 
the  fourth  exceeds  the  second  by  4  V^.  Find  the  first  term 
and  common  ratio. 

42.  In  a  G.  P.  the  sum  of  n  terms  is  S,  and  the  sum  of  27^ 
terms  in  Q8.     Express  the  common  ratio  and  first  term. 

43.  In  a  G.  P.  oi'Zn  -\-l  terms,  whose  first  term  is  5,  the 
Bum  of  the  first  and  last  terms  is  125  greater  than  twice  the 
middle  term.     Find  the  common  ratio. 

44.  The  first  term  of  a  G.  P.  is  2,  and  the  continued 
product  of  the  first  five  terms  is  128.  What  is  the  common 
ratio? 

45.  Find  that  G.  P.  of  which  the  product  of  the  first  and 
second  terms  is  3,  and  that  of  the  third  and  fourth  terms  48. 

46.  A  person  who  each  year  gained  half  as  much  again  as 
he  did  the  year  before,  gained  $2059  in  7  years.  What  was 
his  gain  the  first  year? 


UNIVERSITY 

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